Answers to AP Chemistry Worksheet: Gas Stoichiometry

Answers to AP Chemistry Worksheet: Gas Stoichiometry

1. 1 vol of F2

50ml of Cl2 x ------= 50ml of F2

1 vol of Cl2

2. 2 vol of NH3

10.0 L of N2 x ------= 20 L of NH3

1 vol of N2

3. nRT .009383 mol of H2 x .0821 x 296.5 K

V = ------V = ------

P 1.025atm

V = .233L of H2

4. PN = ¾ of the total pressure because there are 4 moles of gases formed in the products and nitrogen makes up 3 out of the 4 moles and therefore accounts for ¾ of the total pressure.

PN = ¾ x 325atm = 244 atm

5. You must first find the limiting reagent.

Calculations tell me there is .0023 mol of H2 and .0016 mol of O2 since you need twice as much H2 as O2, O2 is the limiting reagent. Based on the balanced chemical equation .0016 moles of O2 will produce .0032 moles of water

nRT .0032mol x .0821 x 303K

V = ------V = ------v = .0780L = 78ml

P 1.02atm

6. Do not worry about this problem there was a mistake in the writing of the problem, SORRY!

7. You must first write a balanced equation for the combustion of octane C8H18

Then determine the number of moles of octane you have – 10.18 moles of octane

From the balanced equation you should be able to determine that 81.44 moles of CO2 would be produced

nRT 81.44 mol x .0821 x 258K

Now, V = ------V= ------V = 42,910 L

P .0402 atm

8. You first must determine the limiting reagent. Calculation tell me I have .1156 moles of NH4Cl and .0540 moles of Ca(OH)2. So, Ca(OH)2 is the limiting reagent.

I then determined from the balance equation that .108 moles of NH3 would be produced

nRT .108 mol x .0821 x 345K

Now, V = ------V = ------V= .862 L of NH3

P 3055atm

9. Ptotal = atm pressure + P vapor pressure

Pressure of CO2 = 752 mm Hg – 16.6 mmHg = 735.4mm Hg/ 760mm Hg = .968 atm

2.00g of Calcium Carbonate is equal to .0199 mole of CaCO3

CaCO3 à CaO + CO2 So, .0199 moles of CO2 should be the theoretical yield

PV .968 atm x .358

n = ------n = ------n = .0145 mol CO2 = actual yield

RT .0821 x 292K

.0145

------x 100 = 72.8%

.0199

2.51 cm 10 mm 1 atm

10. First you must convert 29.38 in of Hg to atm 29.38in X ------X ------X ------= .982atm

1 in 1 cm 760mm

PV .982atm x .072L

n = ------n = ------n = .00291 mol CO2

RT .0821 x 296K

From the equation .00291 moles of CO2 would require .00582 moles of HNO3

63 g of HNO3 1000mg

.0582 moles of HNO3 X ------X ------X ------= 1.41 mg/ml

1 mol of HNO3 1 g 260 ml

11. 2 CU2S + 3O2 à 2Cu2O + 2SO2

First you must determine the limiting reagent you have .0767 moles of Cu2S and .183 moles of O2

To determine the number of moles of O2 I used PV

n =------using the initial conditions.

RT

So, Cu2S is the limiting reagent

nRT .0767 mol of SO2 X .0821 X 373K

V = ------V = ------V = 2.52L of SO2

P .933atm