Chapter 1 - Stoichiometry

Question 1

If we have 8g O2 and unlimited sugar, how much CO2 would be produced?
C12H22O11 + 12O2 => 12CO2 + 11H2O + energy

Answer 1

we have 12 moles of O2 on each side so ratio is 12:12. So just need to convert 8g to moles
n = m/Mr = 8/32 = 1/4 mole of O2
Then 1/4 x 44g/mol = 11g CO2 produced as a result of 8g O2

Question 2

Burning 400mg vitamin C yields 600mg CO2 and 1.632mg of H2O. What are the mg of carbon in 600mg of CO2

Answer 2

(there are 12g of C in CO2)

600 x 12 (C) /44 (CO2) = 1.6mg C in 600mg of CO2 or 400mg of vitamin C

Question 3

What is the percentage by mass of water in the hydrate Na2CO3 10H2O (mass = 286)

Answer 3

So 1 mole of the hydrate has mass of 286 where te 10 moles of water are the hydrate.
n = m/Mr so m = n x Mr
m = 10 x 18g= 180g
% water is 180 / 286 = 62.9%

Question 4

What is the limiting reagent if 78g of Na2O2 were reacted with 29.4g H2O?

Answer 4

78 x 1mol/77g = 0.327 (Na2O2) *limiting
78 x 1mol/18g = 1.6 (H2O) *in excess
the limiting reagent determines the rate of reaction

Question 5

In order to be a combustion reaction, what must be produced?

Answer 5

CO2 and water

Question 6

A system burns 1kg CH4 per day. 5% can’t be regenerated. How many moles of H2 will be lost after a week?

Answer 6

7kg per week where 5% is lost
7 x 0.05 = 0.35kg
H4 is 1/2 of CH4 mass so 350g / 4 = 87mol
or H2 = 43mol will be lost