Mixed Gas Problems:

1. If 690.0 mL of oxygen is collected over water at 26.0 ˚C and 725 mmHg, what is the dry volume of this oxygen sample at 52.0 ˚C and 106.6 kPa?

Given: P₁ = 725 mmHg x (101.3kPa/760mmHg)

= 96.6 kPa

V₁= 690.0 mL, T₁ = 26.0˚C or 299K, T₂ = 52.0˚C or 325 K, P₂ = 106.6 kPa

Required: V₂

Analysis: V₂ = (V₁ x P₁ x T₂) / (P₂ x T₁)

Solve: V₂ = (690.0 mL x 96.6 ~~kPa~~ x 325 K)/ (106.6 ~~kPa~~ x 299 K)

= 679.6 mL

*The dry volume of the oxygen sample at 52.0 ˚C and 106.6 kPa is 679.6 mL.

Evaluate: The answer makes sense because according to Boyle’s Law, pressure is inversely proportional to volume, and since the pressure of the gas increases, the resulting volume decreases.

2. If 400.0 mL of hydrogen gas is collected over water at 18.0˚C and a total pressure of 98.6 kPa, what would be the dry volume of this hydrogen sample at STP?

Given: V₁ = 400.0 mL, T₁ = 18˚C or 291 K, P₁ = 98.6 kPa, STP= 273 K and 101.3 kPa

Required: V₂

Analysis: V₂ = (V₁ x P₁ x T₂) / (P₂ x T₁)

Solve: V₂ = (400mL x 98.6 ~~kPa~~ x 273K)/ (101.3 ~~kPa~~/ 291 K )

= 365.3 mL

*The dry volume of this hydrogen at STP is 365.3 L

Evaluate: According to Boyle’s law, pressure is inversely proportional to volume, while according to Charle’s Law temperature is directly proportional to volume; and since the pressure increases and the temperature decreases, the volume decreases.

3. A 6.12 L sample of xenon gas was collected over water at 2.00 X 10⁵ Pa and 80.0˚C. What would be the gas volume of pure xenon at SATP?

Given: P₁= 2.00 X 10⁵Pa x (1kPa/1 x 10³ Pa)

=200 kPa

V₁= 6.12 L,, T₁ = 80.0˚C or 353 K, SATP= pressure of 101.3kPa and 273 K temperature

Required: V₂

Analysis: V₂ = (V₁ x P₁ x T₂) / (P₂ x T₁)

Solve: V₂ = (6.12 L x 200 kPa x 273 K)/ (101.3kPa x 353 K)

= 9.34 L

*The gas volume of pure xenon at SATP is 9.34 L.

Evaluate: : According to Boyle’s law, pressure is inversely proportional to volume, while according to Charle’s Law temperature is directly proportional to volume; and since the pressure decreases and the temperature increases, the expected outcome of volume would be higher than the original.

4. How many moles of gas are contained in 890.0 mL at 21.0˚C and 750.0 mmHg?

Given: V= 890.0 mL x ( 1L/1000 mL)

= 0.89 L

T= 21.0˚C or 294 K

P= 750 mmHg x (101.3 kPa / 760 mmHg)

=100Kpa

R= 8.31 (L*kPa)/ (K*mol)

Required: mole of the given gas (n)

Analysis: n= PV/RT

Solve: n= (100kPa x 0.89 L) / (8.31 (L*kPa)/ (K*mol) x 294 K)

= 0.04 mol

* There’s 0.04 mol of gas contained in 890.0 mL at 21.0˚C and 750.0 mmHg.

Evaluate:

5. What volume will 20.0 g of Argon gas occupy at STP?

Given: m_argon = 20.0g, STP: T = 273 K, P =101.3 kPa, R= 8.31 (L*kPa)/ (K*mol)

Required: n_argonand volume of gas

Analysis: mol of argon has to be calculated first: n_Ar= 20.0g x (1mol/39.9g)

=0.5 mol of argon

V = nRT/P

= [0.5 mol x 8.31 (L*kPa)/ (K*mol ) x 273 K] / 101.3 kPa

= 11.2 L

*The 20.0 g of Argon will occupy 11.2 L at STP.

6. What volume would 32.0 g NO₂ gas occupy at 3.12 atm and 18.0˚C?

Given: mNO₂ = 32.0 g, T= 18.0˚C or 291 K, R= 8.31 (L*kPa)/ (K*mol)

P= 3.12 atm x (101.3kPa/ 1 atm)

=316.1 kPa

Required: V

Analysis: Calculate mole of NO₂ first,

nNO₂= 32.0 g x [1mol/ 14g + (2x16g)]

=0.7 mol NO₂

V = nRT/P

= (0.7 mol NO₂x 8.31 (L*kPa)/ (K*mol) x 291 K) / 316.1 kPa

= 5.4 L

*The 32.0 g NO₂ gas will occupy 5.4 L at 3.12 atm and 18.0˚C.

7. Calculate the molar mass of a gas if 35.44 g of the gas stored in a 7.50 L tank exerts a pressure of 60.0 atm at 35.5˚C.

Given: m_gas = 35.44g, V= 7.50 L, T= 35.5˚C or 308.5 K, R= 8.31 (L*kPa)/ (K*mol)

P= 60.0 atm x (101.3 kPa/ 1 atm)

= 6078 kPa

Required: n_gas,M_gas

Analyisis: Calculate mole of gas first,

n= PV/ RT

= (6078 kPa x 7.50 L)/ (8.31 (L*kPa)/ (K*mol) x 308.5 K)

= 17.8 mol

Solve: solve for molar mass using mole

M_gas = 35.44 g / 17.8 mol

= 2 g/ mol

*The molar mass of the gas stored in a 7.50 L tank with a pressure of 60.0 atm at 35.5˚C is 2 g / mol.

8. Determine the number of moles of Krypton contained in a 3.25 L tank at 5.88 X 10⁵ Pa and 25.5˚C. If the gas was oxygen instead of krypton, will the answer be the same?

Given: V= 3.25 L, T= 25.5˚C or 298.5 K, R = 8.31 (L*kPa)/ (K*mol)

P = 5.88 x 10⁵ Pa x (1 kPa/ 1x 10³ Pa)

= 588 kPa

Required: n_Kr

Analysis: n= PV/ RT

Solve: n= (588 kPa x 3.25 L)/ (8.31 (L*kPa)/ (K*mol) x 298.5 K)

= 0.77 mol

* The mole of Krypton contained in a 3.25 L tank at 5.88 X 10⁵ Pa and 25.5˚C is 0.77 mol.
Evaluate: If the gas is oxygen instead of krypton , the answer will be the same unless a specific amount of mass is given in the problem.

9. Determine the mass of carbon dioxide in a 450.6 mL flask at 1.80 atm and –50.5˚C. Determine the mass of oxygen that would be present in the same container under the same conditions.

Given: V= 450.6 mL x (1L / 1000mL)

= 0.45 L

P= 1.80 atm x (101.3 kPa/ 1 atm)

= 182.3 kPa

T= -50.5˚C + 273

= 222.5 K

R= 8.31 (L*kPa)/ (K*mol)

Required: n_CO2

Analysis: n= PV/ RT

Solve: n_CO2= (182.3 kPa x 0.45 L) / (8.31 (L*kPa)/ (K*mol) x 222.5 K)

= 0.04 mol

M_CO2 = 0.04 mol x [ 14g + 2(16g)]/ 1 mol

= 18.4 g CO₂

n_O2= (182.3 kPa x 0.45 L) / (8.31 (L*kPa)/ (K*mol) x 222.5 K)

= 0.04 mol

M_O2 = 0.04 mol x (2x16g)/ 1 mol

= 1.28 g O₂

* The mass of CO₂ in a 450.6 mL flask at 1.80 atm and –50.5˚C is 18.4 g CO₂ while the mass of oxygen at the same condition is 1.28 g.

10. At what temperature will 0.654 mol neon gas occupy 12.30 L at 197.5 kPa?

Given: n = 0.654 mol, V= 12.30 L, P= 197.5 kPa, R= 8.31 (L*kPa)/ (K*mol)

Required: T

Analysis: T= PV/ nR

Solve: T= (197.5 kPa x 12.30 L)/ [0.654 mol x 8.31 (L*kPa)/ (K*mol)]

= 447 K

*The 0.654 mol neon gas that occupies 12.30 L at 197.5 kPa will have a temperature of 447 K.

11. A container holds three gases: oxygen, carbon dioxide, and helium. The partial pressures of the three gases are 2.00 atm, 303.9 kPa, and 3040 mm Hg respectively. What is the total pressure of this sample in kPa, atm, and mmHg?

Given: P_oxygen = 2 atm x (101.3 kPa/ 1 atm)

= 202.6 kPa

P_{carbon
dioxide} = 303.9 kPa

P_helium= 3040 mmHgx (101.3 kPa/ 760 mmHg)

= 405.2 kPa

Required: P_Total

Analysis: P_Total = P_oxygen + P_{carbon dioxide} + P_helium

Solve: P_total = 202.6 kPa + 303.9 kPa +405.2 kPa

= 911.7 kPa

911.7kPa x (1atm/ 101.3 kPa)

= 9 atm

911.7kPa x (760 mmHg/ 101.3kPa)

= 6849 mmHg

* The total pressure of this sample is 911.7 kPa or 9 atm or 6489 mmHg.

12. A tank contains 480.0 g of oxygen and 80.0 g of helium at a total pressure of 7.00 mmHg.

What are the partial pressures of oxygen and helium

Gas Stoichiometry:

1. Pentane, C₅H_12(l), burns completely to form carbon dioxide and water.

a) Write the balanced chemical equation for this reaction.

C₅H₁₂+ 8O₂→ 5CO₂ + 6H₂O

b) What volume of O_2(g) at STP is required to produce 70.0 L of CO_2(g) at STP?

Given: V_CO2 = 70.0 L

Required: V_O2, n_O2, n_CO2

Analysis: solve for mole of carbon dioxide and oxygen first, and then calculate the volume.

Solve: n_CO2 =70.0L x ( 1mol/ 22.4 L)

= 3.125 mol CO₂

n_O2 = 3.125 mol CO₂ x (8 mol O₂ / 5 mol CO₂)

=5 mol O₂

V_O2 = 5 mol x (22.4 L / 1 mol)

= 112 L of oxygen

* The volume of oxygen required to produce 70L of CO₂ is 112 L.

c) What mass of H₂O_(l) is made when the combustion of C₅H_12(l) gives 106 L of CO_2(g) at SATP?

Given: V_{carbon
dioxide}

Required: m_water

Analysis: solve for mole of carbon dioxide and water first, and then calculate the mass of water

Solve: : n_CO2 =106.0L x ( 1mol/ 22.4 L)

= 4.7 mol

n_H2O = 4.7 mol CO₂ x (6mol H₂O/ 5 mol CO₂)

= 5.64 mol H₂O

m_water= 5.64mol x [(2 x 1) + 16g / mol]

= 101.52 g of H₂O

*101.52 g of H₂O_(l) is made when the combustion of C₅H_12(l) gives 106 L of CO_2(g) at SATP.

2. Given the equation C₃H₈(g) + O₂(g) → CO₂(g) + H₂O(g)

a) Balance the equation

C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(g)

b) If 50 g of C₃H₈ is burned in excess O₂ , what volume of CO₂ gas can be collected at 30^∧C and 90 kPa?

Given: m_C3H8= 50g, T= 30^∧Cor 303 K, P= 90 kPa

Required: moles of C₃H₈ and CO₂, V_CO2

Analysis: Solve for the mole of C₃H₈ and CO₂first,and then calculate the volume of CO₂

Solve: n_C3H8 = 50 g x [ 1 mol/ (3 x 12) + (8 x 1)g]

= 1.14 mol C₃H₈

n_CO2 = 1.14 mol C₃H₈x (3mol CO₂ / 1mol C₃H₈)

= 3.42 mol CO₂

V_CO2 = nRT/ P

= (3.42 mol x 8.31 (L*kPa)/ (K*mol) x 303 K) / 90kPa

= 95.7 L of CO₂

* 95.7 L of CO₂ gas can be collected at 30^∧C and 90 kPa if 50 g of C₃H₈ is burned in excess O₂.

3. 2 ZnS(s) + 3 O₂(g) → 2 ZnO(s) + 2 SO₂(g)

a) What volume of O₂ at SATP is required for the reaction of 1.46 g of ZnS?

Given: m _ZnS = 1.46 g

Required: mole of O₂ and ZnS; volume of O₂

Analysis: calculate mole of O₂ and ZnS first, then solve for volume of O₂

Solve: n_ZnS= 1.46g x ( 1 mol/ (65.4 + 32.1) g)

= 0.015 mol

n_O2 = 0.015 mol ZnS x (3 mol O₂ / 2 mol Zns)

= 0.023 mol

V_O2= 0.023 mol x (22.4 L / 1 mol)

= 0.52 L O₂

* It requires 0.52 L O₂for the reaction of 1.46 g of ZnS at SATP

c) What volume of SO₂ at SATP will be produced from the reaction in a)?

Given: m _ZnS = 1.46 g

Required: mole of SO₂ and ZnS; volume of O₂

Analysis: calculate mole of SO₂ and ZnS first, then solve for volume of SO₂

Solve: n_ZnS= 1.46g x ( 1 mol/ (65.4 + 32.1) g)

= 0.015 mol

n_SO2 = 0.015 mol ZnS x (2 mol SO₂ / 2 mol Zns)

= 0.015 mol SO₂

V_SO2= 0.015 mol x (22.4 L / 1 mol)

= 0.34 L SO₂

* It requires 0.34 L SO₂for the reaction of 1.46 g of ZnS at SATP.

4. Given the equation (NH₄)₂SO₄(aq) + 2KOH(aq) → 2NH₃(g) + K₂SO₄(aq) + 2H₂O(l)

a) Calculate the volume of ammonia gas, measured at 23^∧C and 64 kPa, that could be produced from 264 g of ammonium sulfate and 280.0 g of potassium hydroxide.

Given: T = 23^∧C or 296 K, P = 64 kPa, mass of (NH₄)₂SO₄= 264 g, mass of KOH = 280.0g, R= 8.31 (L*kPa)/ (K*mol)

Required: n_KOH, n_(NH4)2SO4,n_{ammonia gas,} V_{ammonia gas}

Analysis: calculate the moles of KOH, (NH₄)₂SO₄, and NH₃then solve for volume of ammonia

Solve: n_KOH = 280 g x [1mol /(39.1 + 16 + 1)g]

=5 mol KOH

n_{NH3 =}5 mol KOH x (2 mol NH₃/ 2 mol KOH)

= 5 mol NH₃

V_NH3 = nRT/P

= (5 mol x 8.31 (L*kPa)/ (K*mol) x 296 K) / 64 kPa

= 192 L

n_(NH4)2SO4 = 264 g x [1mol/ ( (2x14)+ (8 x 2) + 32.1 + ( 4 x 16)]

= 1.88 mol

n_NH3= 1.88 mol (NH₄)₂SO₄x ( 2 mol KOH/ 1 mol (NH₄)₂SO₄)

= 3.76 mol NH₃

V_NH3 = nRT/P

= (3.76 mol x 8.31 (L*kPa)/ (K*mol) x 296 K) / 64 kPa

= 144.5 L

* At 23^∧C and 64 kPa, there will be 192 L of NH₃ produced in 280 g KOH and 144.5 L of NH₃ produced in 264 g of (NH₄)₂SO_4.

5. Given the equation 2KClO₃(s) → 2KCl (s) + 3O₂(g)

What volume of a gas can be produced by the decomposition of 122.6 g of potassium chlorate measured under the following conditions?

a) at STP

Given: m _KClO3= 122. 6g , P = 101.3 kPa, T = 273 K, R= 8.31 (L*kPa)/ (K*mol)

Required: n_KClO3, n_O2, V_O2

Analysis: Solve the mole of KClO₃ and O₂ first, then calculate the volume

Solve: n_KClO3= 122. 6g x (1mol/ [39.1 + 35.5 + (3 x 16)]

= 1 mol

n_O2 = 1 mol KClO₃ x (3 mol O₂/2 mol KClO₃)

= 1.5 mol O₂

V_O2 = nRT/ P

=(1.5 mol x 8.31 (L*kPa)/ (K*mol) x 273 K) / 101.3 kPa

= 33.6 L

* At STP, the volume of a gas can be produced by the decomposition of 122.6 g of potassium chlorate is 33.6 L.

b) at SATP

Given: m _KClO3= 122. 6g , P = 101.3 kPa, T = 298 K, R= 8.31 (L*kPa)/ (K*mol)

Required: n_KClO3, n_O2, V_O2

Analysis: Solve the mole of KClO₃ and O₂ first, then calculate the volume

Solve: n_KClO3= 122. 6g x (1mol/ [39.1 + 35.5 + (3 x 16)]

= 1 mol

n_O2 = 1 mol KClO₃ x (3 mol O₂/2 mol KClO₃)

= 1.5 mol O₂

V_O2 = nRT/ P

=(1.5 mol x 8.31 (L*kPa)/ (K*mol) x 298 K) / 101.3 kPa

= 36.7 L

* At STP, the volume of a gas can be produced by the decomposition of 122.6 g of potassium chlorate is 36.7 L.

EVALUATE: The volume at STP condition is smaller than at SATP since the temperature at SATP is greater than STP, and according to Charle's Law, temperature is directly proportional to volume, so the greater volume for SATP is acceptable.

Chemistry Lab Group 5 A.M GVSS

Friday, June 15, 2012

Mixed Gas Problems:

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