Final Exam

Question 1

Magnesium has three stable isotopes, and Chlorine has two stable isotopes. How many discrete masses can a single molecule of Magnesium Chloride (MgCl2) have?

Answer 1

A single molecule of MgCl2 can have one of nine discrete masses. Each isotope of Magnesium can have one of three combinations of Chlorine atoms.

(Mg 24, Mg 25, and Mg 26) each can have one of the following combinations (Cl 35, Cl 35 – Cl 37, Cl 37 – Cl 35, Cl 37).

Remember, this is NOT asking about the number of possible combinations between the Magnesium and Chlorine isotopes. It is only asking about a SINGLE molecule.

Question 2

Balance:

C6H12 + O2 –> CO2 + H2O

Answer 2

C6H12 + 9 O2 –> 6 CO2 + 6 H2O

Question 3

Balance:

BF3 + NaH –> B2H6 + NaF

Answer 3

2 BF3 + 6 NaH –> B2H6 + 6 NaF

Question 4

Balance:

LiAlH4 + AlCl3 –> AlH3 + LiCl

Answer 4

3LiAlH4 + AlCl3 –> 4 AlH3 + 3 LiCl

Question 5

Balance:

GeCl4 + SO3 –> Ge(SO4)2 + S2O5Cl2

Answer 5

GeCl4 + 6 SO3 –> Ge(SO4)2 + 2 S2O5Cl2

Question 6

Balance:

HNO3 + H2O + As2O3 –> N2O3 + H3AsO4

Answer 6

2 HNO3 + 2 H2O + As2O3 –> N2O3 + 2H3AsO4

Question 7

To perform a stoichiometric equation, what variables must you know?

Answer 7

The mole ratios of the compounds and their molar masses.

Question 8

Stoichiometric Equation:

Suppose you have the following:
N2 + 3H2 –> 2NH3

How many grams of ammonia can be made from 3.00 g Hydrogen?

Answer 8

(Moles of Given) x (Ratio (Wanted / Given)) x ( g / mol of Wanted)

3.00 g H2 (2 mol NH3 / 3 mol H2) (17.03 g/mol)

Answer = 16.9 g NH3

Question 9

Limiting Reagent Equation:

Suppose you have the following:
P4O10 + 6 PCl5 –> 10 POCl3

If you have 5.50 g P4O10 and 23.00 g PCl5, which is the limiting reagent?

Answer 9

Treat it as two separate equations. You have two reactants and one product, so you need to calculate for each reactant and its theoretical output individually.

(Moles of Given Reactant) x (Ratio (Wanted / Given)) x ( g / mol of Wanted)

5.50 g P4O10 ( mol / 283.88 g ) (10 mol POCl3 / mol P4O10) (153.22 g/mol)

Answer = 29.69 g POCl3

Question 10

How do you calculate percent yield?

Answer 10

Percent yield = actual yield / theoretical yield

where theoretical yield is calculated in the stoichiometric or limiting reagent equation.

Question 11

Concentration:

2.50g of NaBr is dissolved in water to make a 35.0 mL solution. What is the concentration of Na+ ions in the solution?

Answer 11

Molarity = (moles of solute) / (Liters of solution)

2. 50g NaBr * (mol / 102.89 g) = 0.0243 mol NaBr
3. 0243 mol NaBr * ( 1 mol Na / 1 mol Br) = 0.0243 mol NaBr
4. 0243 mol NaBr / 0.0350 L = 1.323 M Na+

Question 12

Orbital strength depends on what?

Answer 12

Strong bonds require that orbitals be of COMPARABLE SIZE and COMPARABLE ENERGY.