Test 3

Test week three: Percent composition Amount of reactants and products, Limiting reagents

1) Phosphorus oxychloride is a starting compound for preparing substances used as flame retardants in plastics. If an 8.53 mg sample of phosphorus oxychloride contains 1.72 mg of phosphorus. What is the mass percent of phosphorus in the compound?

2) Two compounds have identical percent composition, 88.83% C and 11.17% Hydrogen.

a. What is the empirical formula corresponding to this composition?

b. One of the compounds has a molecular weight of 54.08 amu and the other has a molecular weight of 108.16 amu, what are the empirical formulas of both compounds?

3) Balance the following chemical equations:

a) Na + H₂O -> NaOH + H₂

b) P₄ + H₂ -> PH₃

4) Write and balance the chemical equation for the combustion of gaseous ammonia. NH₃ is combusted by reacting with oxygen in the air to form gaseous nitrogen and water.

5)Given the chemical reaction for the formation of methanol: CO + 2H₂ -> CH₃OH;

41.5 grams of CO and 8.25 g H₂, How many grams of methanol would be produced in a complete reaction? Which reactant remains unconsumed and how many grams remain?

Answers

1) (1.72 mg phosphorus/8.53 mg phosphorus oxychloride)x100 = 20.2%

2)

a) Assume you have 100 grams of the compound thus you have 88.83 g C and 11.17 g H. Calculate the number of moles of each:

88.83 g C x (1 mol/12.01 g) = 7.40 mols

11.17 g H x (1 mol/1.008 g) = 11.08 mols

C_7.40H_11.08 divide through by 7.40 mol and you get CH_1.5 multiply by two and you get C₂H₃

b) Next given the molecular weights 54.08 amu and 108.16 amu, determine the molecular formula from the two empirical formulas.

C₂H₃ has a molecular weight of 27.044, 54.08 amu is roughly double therefore its molecular formula is C₄H₆. 108.16 amu is roughly quadruple therefore its molecular formula is C₈H₁₂.

3) a) 2Na + 2H₂O -> 2NaOH + H₂

b) P₄ + 6H₂ -> 4PH₃

4) 4NH₃ + 3O₂ -> 2N₂ + 6H₂O

5) Determine the limiting reagent

41.5 g CO (1mol CO/28.01 g CO) x (1 mol CH₃OH/1 mol CO) = 1.48 mol CH₃OH

8.25 g H₂ (1mol H₂/2.016 g H₂)x(1 mol CH₃OH/ 2 mol H₂) = 2.04 mol CH₃OH

Convert mols of CH₃OH to grams

1.48 mol CH₃OH x (32.04 g CH₃OH/1mol CH₃OH) = 47.4 g CH₃OH

Reactant unconsumed

1.48 mol CH₃OH x (2 mol H₂/1mol CH₃OH)x(2.016 g H₂/1mol H₂)

= 5.97 g H₂ react therefore 2.28 grams remain