CHEM CALC

Question 1

Wine making involves a series of very complex reactions most of
which are performed by microorganisms. The starting
concentration of sugars determines the final alcohol content and
sweetness of the wine. The specific gravity of the starting stock
is therefore adjusted to achieve desired quality of wine. A starting
stock solution has a specific gravity of 1.075 and contains 12.7
wt% sugar. If all the sugar is assumed to be C12H22O11, determine

1. kg sugar / kg H2O
   a. 0.286 c. 0.145
   b. 0.039 d. 0.327
2. lb solution / ft3 solution
   a. 67.1 c. 50.5
   b. 39.2 d. 48.6
3. g sugar / L solution
   a. 202 c. 295
   b. 136 d. 144

Answer 1

1. C. 0.145
2. A. 67.1
3. B. 136

Question 2

Two aqueous sulfuric acid solutions containing 20 wt% H2SO4
(SG=1.139) and 60 wt% H2SO4 (SG=1.498) are mixed to form a
4 M solution (SG=1.213).

1. Calculate the mass fraction of sulfuric acid in the product solution.
   a. 0.105 c. 0.323
   b. 0.667 d. 0.548
2. Calculate the feed ratio (liters 20% solution per liter 60% solution)
   a. 2.96 c. 1.75
   b. 3.08 d. 4.51
3. What feed rate of the 60% solution (L/hr) would be required to
   produce 1250 kg/h of the product.
   a. 986 L/h c. 110 L/h
   b. 257 L/h d. 462 L/h

Answer 2

1. C. 0.323
2. A. 2.96
3. B. 257 L/h

Question 3

If 1 L of ethyl alcohol is mixed with 1 L of water at 200C.

1. How many kilograms of solution result?
   a. 1.85 kg c. 1.79 kg
   b. 2.06 kg d. 2.11 kg
2. How many liters?
   a. 1.93 L c. 2.00 L
   b. 1.88 L d. 2.15 L

Answer 3

1. C. 1.79 kg
2. A. 1.93 L

Question 4

Air is bubbled through a drum of liquid hexane at a rate of 0.10
kmol/min. The gas stream leaving the drum contains 10 mol%
hexane vapor. Air may be considered insoluble in liquid hexane.
Use an integral balance to estimate the time required to vaporize
10 m3 of the liquid.

a. 4200 min c. 7700 min
b. 6900 min d. 8100 min

Answer 4

B. 6900 min

Question 5

Urea, phosphoric acid and potassium chloride are mixed together
to obtain a mixed fertilizer having NPK content 10:26:26 as %N,
%P2O5 and %K2O by weight, balance being the weight of filler
materials. Calculate the amount of potassium chloride to be
mixed to get 1000 kg of mixed fertilizer.

a. 214.4 kg c. 411.5 kg
b. 359 kg d. 505 kg

Answer 5

C. 411.5 kg

Question 6

A supplier shipped a material for chicken feed with the following
analytical report: protein – 22%, triglyceride – 8%, moisture –
10%. During transport moisture was picked up by the material;
the later analytical report showed 14% moisture. Based on the
latest report, recalculate the % triglyceride.

a. 21.02% c. 7.64%
b. 39.33% d. 14.65%

Answer 6

C. 7.64%

Question 7

An evaporator is fed continuously with 25 MT/h of a solution
consisting of 10% NaOH, 10% NaCl, and 80% H2O. During
evaporation, water is boiled off, and salt precipitates as crystals,
which are settled and removed from the remaining liquor. The
concentrated liquor leaving the evaporator contains 50% NaOH,
2% NaCl and 48% H2O.

1. Calculate the kg of water evaporated per hour
   a. 19300 c. 17600
   b. 18500 d. 16200
2. Calculate the kg salt precipitated per hour
   a. 2400 c. 1100
   b. 3500 d. 4700
3. Calculate the kg of concentrated liquor produced per hour
   a. 4000 c. 6000
   b. 3000 d. 5000

Answer 7

1. C. 17600
2. A. 2400
3. D. 5000

Question 8

An experimental oil is used to distil 6000 kg/hr of Athabasta tar sand which analyzes 30% “heavy oil”, “light oil”, and non-volatile matter. If the distillate contains 12% heavy oil and this constitutes 25% recovery of the component the weight in kg of the distillate obtained is.

Answer 8

3750 kg/hr

Question 9

Ten kilograms each of salts A and B are dissolved in 50 kg of
water. The solubility of A in water is 1 kg per kg water and that of
B is 0.5 kg per kg of water. The solution is heated to evaporate water. When the concentration of the solution reaches the
saturation concentration of any salt, further evaporation results in
the crystallization of that salt. If 70% of the water that is originally
present in the solution is driven off by evaporation, determine the
following:

1. The weight of the final solution
   a. 46.5 c. 11.5
   b. 28.5 d. 32.5
2. The weight of the crystals formed
   a. 2.5 c. 0
   b. 5.5 d.1.5

Answer 9

1. D. 32.5
2. A. 2.5

Question 10

1200 lb of barium nitrate is dissolved in sufficient water to form a
saturated solution at 900C. Assuming that 5% of the weight of the
original solution is lost through evaporation calculate the crop of
crystals obtained when cooled at 200C. Solubility data for barium
nitrate at 900C is 30.6% wt and 8.6 wt% at 200C.

a. 952.4 lbs c. 962.4 lbs
b. 970 lbs d. 980 lbs

Answer 10

C. 962.4 lbs

Question 11

What is the theoretical yield of crystals which may be obtained by
cooling a solution containing 1000 kg of sodium sulfate (mw = 142
kg/kmol) in 5000 kg water to 283 K? The solubility of sodium
sulfate at 283 K is 9 kg anhydrous salt / 100 kg water and the
deposited crystals will consist of decahydrate. It may be assumed
that 2% of the water will be lost by evaporation during cooling.

a. 4469.17 kg c. 6000 kg
b. 250 kg d. 1430.83 kg

Answer 11

D. 1430.83 kg

Question 12

A solution of NH4Cl is saturated at 700C. Calculate the
temperature to which this solution must be cooled in order to
crystallize out of 45% of the NH4Cl. The solubility of NH4Cl in
water are:
Temperature, 0C Solubility,
g/100g H2O
70 60.2
10 33.3
0 29.4

a. 8.6 c. 9.5
b. 5.7 d. 7.2

Answer 12

C. 9.5

Question 13

In the azeotropic distillation of an ethanol–water solution, a feed
mixture containing 95.6% alcohol is mixed with pure benzene and
distilled. The benzene forms a ternary azeotrope with alcohol–
water with a composition of 74.1% benzene, 7.4% water and
18.5% alcohol, which is distilled over as the overhead product.
Absolute alcohol is obtained as the residue product. Determine
the quantity of benzene required for producing 100 kg of absolute
alcohol.

a. 48 kg c. 66 kg
b. 74 kg d. 52 kg

Answer 13

D. 52 kg

Question 14

Oil is extracted from seeds by leaching with organic solvents.
Soybean seeds containing 20% oil, 65% inert solids and 15%
water are leached with hexane and after extraction the solid
residue is removed from the solution of oil in hexane. The residue
analyzed 1.0% oil, 88% inert cake and 11% water. What percent
of oil is recovered?

a. 90% c. 96%
b. 82% d. 87%

Answer 14

C. 96%

Question 15

Acetone is recovered from an acetone-air mixture containing 25%
(volume) acetone by scrubbing with water. Assuming that air is
insoluble in water, determine the percent of acetone in the entering gas that is absorbed if the gas leaving the scrubber
analyzes 5% acetone.

a. 96% c. 65%
b. 84% d. 73%

Answer 15

B. 84%

Question 16

One hundred pounds of solution containing 80 wt% Na2SO4 must
be obtained by evaporating a dilute solution using a double effect
evaporator. If the evaporator from the 2nd effect is 60% of that
coming from the first effect and the concentrate of the first effect
contains 1 lb Na2SO4 per pound of water. Calculate for the
composition of the feed.

a. 69% Na2SO4 c. 45% Na2SO4
b. 55% Na2SO4 d. 31% Na2SO4

Answer 16

D. 31% Na2SO4

Question 17

One drier will dry a material from 45% moisture (wb) to 20%
moisture (db) from here the material enters another drier where
the H2O content is further reduced to give a final product weighing
1000 kg. If the total evaporation from both driers is 800 kg, the
moisture of the final product is

a. 5% c. 2%
b. 3% d. 1%

Answer 17

D. 1%

Question 18

An aqueous solution containing 15% NaOH and 0.5% NaCl is
concentrated at a rate of 100 kg/min in an evaporator. The
concentrated solution is then mixed with 2000 kg of aqueous
NaOH solution in a mixer. At the end of one hour a sample is
collected from the mixer and analyzed. The analysis shows 40%
NaOH and 0.8574% NaCl. Calculate the following:

1. The concentration of the original solution in the receiver
   a. 32% NaOH c. 40% NaOH
   b. 25% NaOH d. 56% NaOH
2. The mass of water (in kilograms) evaporated in one hour
   a. 3200 kg c. 4500 kg
   b. 5100 kg d. 6600 kg

Answer 18

1. B. 25% NaOH
2. C. 4500 kg

Question 19

For the preparation of potassium nitrate, 10000 kg/h of a 20%
KNO3 solution is mixed with a recycle stream and sent to an
evaporator. The rate of evaporation is 1.5 times the rate of
introduction of recycle stream. The concentrated solution leaving
the evaporator contains 50% KNO3. This is admitted to the
crystallizer which yields crystals containing 5% water. At the
crystallization temperature the solubility is 50 kg/100 kg of water.
The major part of the mother liquor leaving the crystallizer is
returned to the crystallizer as recycle. Calculate the following:

1. The concentration of KNO3 in the stream entering the evaporator
   a. 33% c. 16%
   b. 25% d. 40%
2. The flow rate of recycle stream in kg/h
   a. 5143 c. 1354
   b. 4531 d. 3415
3. The rate of production of crystals
   a. 3925 c. 1588
   b. 2007 d. 4610

Answer 19

1. B. 25%
2. A. 5143
3. B. 2007

Question 20

Fresh air containing 4 mol% water vapor is to be cooled and
dehumidified to a water content of 1.70 mol% H2O. A stream of
fresh air is combined with a recycle stream of previously
dehumidified air and passed through the cooler. The blended stream entering the unit contains 2.3 mol% H2O. In the air
conditioner, some of the water in the feed stream is condensed
and removed as liquid. A fraction of the dehumidified air leaving
the cooler is recycled and the remainder is delivered to a room.
Taking 100 mol of dehumidified air delivered to the room as a
basis of calculation, calculate the following:

1. The moles of fresh feed
   a. 115 mol c. 102 mol
   b. 126 mol d. 133 mol
2. The moles of water condensed
   a. 0.8 mol c. 2.4 mol
   b. 1.0 mol d. 3.3 mol
3. The moles of dehumidified air recycled
   a. 270 mol c. 280 mol
   b. 290 mol d. 260 mol

Answer 20

1. C. 102 mol
2. C. 2.4 mol
3. B. 290 mol

Question 21

An equimolar liquid mixture of benzene and toluene is separated
into two product streams by distillation. A process flowchart and
a somewhat oversimplified description of what happens in the
process follow:

Inside the column a liquid stream flows downward and a vapor
stream rises. At each point in the column some of the liquid
vaporizes and some of the vapor condenses. The vapor leaving
the top of the column, which contains 97 mole% benzene, is
completely condensed and split into two equal fractions: one is
taken off as the overhead product stream, and the other (the
reflux) is recycled to the top of the column. The overhead product
stream contains 89.2% of the benzene fed to the column. The
liquid leaving the bottom of the column is fed to a partial reboiler
in which 45% of it is vaporized. The vapor generated in the
reboiler (the boil-up) is recycled to become the rising vapor
stream in the column, and the residual reboiler liquid is taken off
as the bottom product stream. The compositions of the streams
leaving the reboiler are governed by the relation:
yB (1-yB)
xB (1-xB) = 2.25

where and are the mole fractions of benzene in the vapor and
liquid streams, respectively. Take a basis of 100 mol fed to the
column, determine the following:

1. The molar amounts of overhead and bottom products
   a. 46 mol and 54 mol c. 33 mol and 67 mol
   b. 49 mol and 51 mol d. 38 mol and 62 mol
2. The mole fraction of benzene in the bottom product
   a. 0.2 c. 0.4
   b. 0.3 d. 0.1
3. The percentage recovery of toluene in the bottoms product
   a. 92% c. 95%
   b. 97% d. 90%

Answer 21

1. A. 46 mol and 54 mol
2. D. 0.1
3. B. 97%

Question 22

Fresh orange juice contains 12% (by weight) solids and the rest water, 90% of the fresh juice is sent to an evaporator to remove water and subsequently mixed with the remaining 10% of the fresh juice. The resultant product contains 40% solids. The kg of water removed from 1 kg fresh juice is.

Answer 22

0.70 kg

Question 23

Fresh orange juice contains 12.0 wt% solids and the balance
water, and the concentrated orange juice contains 42.0 wt%
solids. Initially a single evaporation process was used for the concentration, but volatile constituents of the juice escaped with
the water, leaving the concentrate with a flat taste. The current
process overcomes this problem by bypassing the evaporator
with a fraction of the fresh juice. The juice is concentrated to 58
wt% solids, and the evaporator product stream is mixed with the
bypassed fresh juice to achieve the desired final concentration.
Calculate the following:

1. The amount of product (42% concentrate) produced per 100 kg
   fresh juice fed to the process.
   a. 35.3 c. 28.6
   b. 40.5 d. 19.2
2. The fraction of the feed that bypasses the evaporator.
   a. 0.032 c. 0.150
   b. 0.099 d. 0.250

Answer 23

1. C. 28.6
2. B. 0.099

Question 24

A limestone analysis: CaCO3 – 92.89%, MgCO3 – 5.41% and
unreactive – 1.70%. By heating the limestone, you recover oxides
that together known as lime:

1. How many pounds of calcium oxide can be made from 1 ton of
   limestone?
   a. 1041 lb c. 2348 lb
   b. 3567 lb d. 4020 lb
2. How many pounds of CO2 can be recovered per lb of limestone?
   a. 0.33 c. 0.22
   b. 0.11 d. 0.44
3. How many pounds of limestone are needed to make 1 ton of
   lime?
   a. 3100 c. 3550
   b. 3890 d. 3760

Answer 24

1. A. 1041 lb
2. D. 0.44
3. C. 3550

Question 25

The synthesis of ammonia proceeds according to the following
reaction:
N2 + 3 H2 —> 2 NH3

In a given plant, 4202 lb of nitrogen and 1046 lb of hydrogen are
fed to the synthesis reactor per hour. Production of pure ammonia
from this reactor is 3060 lb per hour.

1. What is the limiting reactant?
   a. N2 c. both (a) and (b)
   b. H2 d. NH3
2. What is the percent excess reactant?
   a. 13% c. 17%
   b. 16% d. 19%
3. What is the percent conversion obtained based on the limiting
   reactant?
   a. 50% c. 70%
   b. 60% d. 80%

Answer 25

1. A. N2
2. B. 16%
3. B. 60%

Question 26

A barytes composed of 100 percent BaSO4 is fused with carbon
in the form of coke containing 6% ash (which is infusible). The
composition of the fusion mass is

BaSO4 11.1%
BaS 72.8%
C 13.9%
Ash 2.2%
Reaction: BaSO4 + 4 C à BaS + 4 CO

1. Find the excess reactant.
   a. C c. CO
   b. BaSO4 d. BaS
2. Find the percent excess reactant.
   a. 44.8% c. 50.6%
   b. 38.9% d. 66.7%
3. What is the degree of completion of the reaction?
   a. 0.6 c. 0.8
   b. 0.7 d. 0.9

Answer 26

1. A. C
2. C. 50.6%
3. D. 0.9

Question 27

In the Deacon process for the manufacture of chlorine, HCl and
O2 react to form Cl2 and H2O. Sufficient air is fed to provide 35%
excess oxygen and the fractional conversion of HCl is 85%.
Calculate the mole fraction of water in the product stream.

a. 0.18 c. 0.10
b. 0.12 d. 0.15

Answer 27

A. 0.18

Question 28

Phenol (94) can be manufactured by reacting chlorobenzene
(112.45) with sodium hydroxide (40). To produce 1000 kg of
phenol, 1200 kg of sodium hydroxide and 1,320 kg of chlorobenzene are used. What is the maximum conversion of the
excess reactant?

a. 29.66% c. 31.05%
b. 35.47% d. 39.13%

Answer 28

D. 39.13 %

Question 29

Acrylonitrile is produced in the reaction of propylene, ammonia
and oxygen: C3H6 + NH3 + 3/2 O2 à C3H3N + 3 H2O
. The feed contains 10 mole % propylene, 12 mole% ammonia,
and 78% air. A fractional conversion of 30% of the limiting
reactant is achieved. Taking 100 mol of feed as basis, determine
the following:

1. The percentage by which of the reactants in excess
   a. 24%, 6.2% c. 25%, 8.5%
   b. 28%, 7.5% d. 20%, 9.2%
2. The mole% of water in the product gas
   a. 6% c. 5%
   b. 7% d. 9%

Answer 29

1. D. 20%, 9.2%
2. D. 9%

Question 30

Given the reaction of ethyl tetrabromide with zinc dust produces
C2H2 and ZnBr2. Based on the C2H2Br4, on one pass through the
reactor the conversion is 80%, and the unreacted C2H2Br4 is
recycled. On the basis of 1000 kg/hr of C2H2Br4 fed to the reactor
per hour, calculate

1. The rate of recycle in lb/hr
   a. 150 kg c. 350 kg
   b. 250 kg d. 450 kg
2. The amount of Zn that has to be added per hour if Zn is to be 20%
   in excess.
   a. 233 kg c. 454 kg
   b. 116 kg d. 378 kg
3. The mole ratio of ZnBr2 to C2H2 in the products
   a. 1.5 c. 2.0
   b. 1.0 d. 2.5

Answer 30

1. B. 250 kg
2. C. 454 kg
3. C. 2

Question 31

TiCl4 can be formed by reacting titanium dioxide (TiO2) with hydrochloric acid. TiO2 is available as an ore containing 78 %
TiO2 and 22 % inerts. The HCl is available as 45 wt% solution
(the balance is water). The per pass conversion of TiO2 is 75 %. The HCl is fed into the reactor in 20 % excess based on the
reaction. Pure unreacted TiO2is recycled back to mix with the
TiO2 feed.

TiO2+ 4 HCl –> TiCl4 + 2H2O

For 1 kg of TiCl4 produced, determine:

1. The kg of TiO2 ore fed.
   a. 0.26 c. 2.05
   b. 0.54 d. 1.33
2. The kg of 45 wt % HCl solution fed.
   a. 0.26 c. 2.05
   b. 0.54 d. 1.33
3. The ratio of recycle stream to fresh TiO2 ore (in kg).
   a. 0.26 c. 2.05
   b. 0.54 d. 1.33

Answer 31

1. B. 0.54
2. C. 2.05
3. A. 0.26

Question 32

Propane is dehydrogenated to form propylene in a catalytic
reactor. The process is to be designed for a 95% overall
conversion of propane. The reaction products are separated into
two streams: the first, which contains H2, C3H6, and 0.555% of the
propane that leaves the reactor, is taken off as a product; the
second stream, which contains the balance of the unreacted
propane and 5% of the propylene in the first stream, is recycled
to the reactor.

1. Calculate the ratio of moles recycled / mole fresh feed
   a. 1.2 c. 2.5
   b. 6.8 d. 9.0
2. What is the single-pass conversion?
   a. 9.6% c. 25.0%
   b. 13.5% d. 38.2%

Answer 32

1. D. 0.90
2. A. 9.6 %

Question 33

Butane is burned with air. No carbon monoxide is present in the combustion products. Calculate the molar composition of water
in the product gas for each of the following cases:

1. Theoretical air supplied, 100% conversion of butane
   a. 12.0% c. 3.3%
   b. 14.9% d. 10.7%
2. 20% excess air, 100% conversion of butane
   a. 12.6% c. 15.2%
   b. 8.8% d. 10.1%
3. 20% excess air, 90% conversion of butane
   a. 9.1% c. 11.4%
   b. 13.6% d. 7.5%

Answer 33

1. B. 14.9 %
2. A. 12.6%
3. C. 11.4%

Question 34

A fuel containing 75% ethane and 25% propane is burned with
dry air. All the H2 burns to H2O and the CO2 to CO ratio is 10:1.
Fifteen% excess air is supplied. Calculate:

1. Percent CO in the Orsat analysis of stack gas
   a. 1.04% c. 5.33%
   b. 2.18% d. 4.27%
2. Percent CO in the complete analysis of stack gas
   a. 0.65% c. 0.89%
   b. 0.33% d. 0.99%

Answer 34

1. A. 1.04%
2. C. 0.89%

Question 35

Orsat analysis of the stack gas from the combustion of a gaseous mixture of acetylene and methane shows 9.82% CO2, 1.37%CO,
0.69% H2, 5.43% O2 and 82.69% N2. Determine:

1. The % excess air
   a. 15% c 10%
   b. 20% d. 25%
2. The % mol CH4 in the gaseous fuel
   a. 50% c. 40%
   b. 30% d. 60%

Answer 35

1. D. 25%
2. D. 60%

Question 36

A pure saturated hydrocarbon is burnt with excess air. Orsat
analysis of the stack gas shows 7.9% CO2, 1.18% CO, 0.24% H2,
5.25% O2, and 85.43% N2. Air is substantially dry. The stack
gases leave at 750 mmHg pressure. Calculate:

1. The % excess air
   a. 25% c 18%
   b. 34% d. 44%
2. The formula of the hydrocarbon
   a. CH4 c. C3H8
   b. C2H6 d. C4H10

Answer 36

1. A. 25%
2. A. CH4

Question 37

The burning of pure butane with excess air gives a stack gas
which analyzes 11.55% CO2 on a dry basis. Assuming complete
combustion, calculate the % excess air

a. 16% c. 11%
b. 20% d. 23%

Answer 37

B. 20%

Question 38

The octane number of a gasoline was determined using a mixture of isooctane and n-heptane with the same knocking tendency as
the gasoline If the iso-octane - heptane mixture is burned in 30%
excess air, with a product gas molal ratio of CO2 to CO of 5:2 and
H2 to CO of 1:1, what is the % CO in the Orsat analysis? Assume
octane number of 85. Density of iso-octane is 0.6918 and n-
heptane is 0.684 g/ml.

a. 3.03% c. 2.95%
b. 1.56% d. 4.47%

Answer 38

C. 2.95%

Question 39

If the test study on the combustion of the cetane-methyl
naphthalene mixture gave a product gas analyzing 7.14% CO2,
4.28% CO, 8.24% O2 and 80.34% N2, what is the cetane number
of the diesel. Density of cetane is 0.7751 and methyl naphthalene
is 1.025 g/ml.

a. 28 c. 53
b. 36 d. 45

Answer 39

D. 45

Question 40

A furnace is fired with sub bituminous B coal containing 10.3%
moisture, 34% VCM and 7.7% ash. It is also known to contain
1.2% N and 1.57% S. Its calorific value is 22 MJ/kg. Calculate its:

1. % Fixed carbon
   a. 55% c. 48%
   b. 37% d. 62%
2. % Combined water
   a. 22.5% c. 18.5%
   b. 10.7% d. 35.9%
3. % Oxygen
   a. 29% c. 38%
   b. 40% d. 15%
4. % Combined water in VCM
   a. 55% c. 66%
   b. 44% d. 33%
5. Calorific Value of VCM.
   a. 12 MJ/kg c. 15 MJ/kg
   b. 17 MJ/kg d. 10 MJ/kg

Answer 40

1. C. 48%
2. A. 22.5%
3. A. 29%
4. C. 66%
5. B. 17 MJ/kg

Question 41

A furnace is fired with high volatile A bituminous coal whose
ultimate analysis shows 75.2% C, 5.19% H, 8.72% O, 1.5% N,
7.8% ash and 1.6% S. 60% excess air is supplied. Assume CO
to CO2 ratio of 0.175. The stack gas leaves at 3000C, 740 torrs.
Calculate:

1. The % water in the complete analysis of the stack gas if air is supplied at 280C, 100 kPa and essentially dry.
   a. 2.06% c. 3.12%
   b. 5.90% d. 4.47%
2. The calorific value of the coal.
   a. 28 MJ/kg c. 31 MJ/kg
   b. 25 MJ/kg d. 39 MJ/kg

Answer 41

1. D. 4.47%
2. C. 31 MJ/kg

Question 42

A high volatile B bituminous coal analyzing 22% VCM, 64% FC,
4% M, 1.4% N and 1.6% S has a calorific value of 32.5 MJ/kg. It
is burned in excess air supplied essentially dry at 280C and 1 atm.
The stack gases leave at 250C, 740 mmHg and contain 8.37%
CO2, 4.19% CO and 2.51% H2. Calculate:

1. The % excess O2
   a. 38% c. 55%
   b. 40% d. 62%
2. The % N2 in the complete Orsat analysis of the stack gas
   a. 75% c. 76%
   b. 89% d. 80%

Answer 42

1. B. 40%
2. C. 76%

Question 43

A furnace burns coal with the following analysis: M – 4.1%, VCM
– 24%, FC – 63.0%, N – 1.20%, S – 1.80%, Ash – 8.90%. The
refuse analyzed 4.8% VCM, 12.6% FC, 82.6% ash and a calorific
value of 32 MJ/kg. Calculate the percentage of gross calorific
value lost in the refuse.

a. 1.38% c. 0.90%
b. 2.16% d. 3.05%

Answer 43

B. 2.16%