Question

1Q)Calculate the number of liters of oxygen that are needed to react with 410. mL of CH₄ at 70.0 °C and 730 mm Hg in a complete combustion reaction.

2Q)Given a sample of gas that exerts 5.54 atm of pressure in a 10.0-L container at 30.0 °C, calculate the new temperature if the pressure is lowered to 2.95 atm.

3Q)A newly discovered gas has a density of 2.30 g/L at 23.0 °C and 733 mm Hg. What is the molar mass of the gas?

Answer 1

Q1)

0.820L

Explanation

Ideal gas equation is

PV = nRT

P= Pressure, 730mmHg = 0.9605atm

V = volume, 410 = 0.410L

R = gas constant, 0.082057(L atm /mol K)

T = Temperature, 70.0°C = 343.15K

n = no of moles

n = PV/RT

= 0.9605atm × 0.410L /( (0.082057L atm/mol K) × 343.15K)

= 0.01399

No of moles of CH4 = 0.01399

now, consider the combination reaction of methane

CH4(g) + 2O2(g) - - - - > CO2(g) + 2H2O(l)

Stoichiometrically, 2moles of O2 required to react with 1mole of CH4

Therefore

No of of moles of O2 required = 2× 0.01399mol = 0.02798

Now, apply ideal gas equation to calculate volume of oxygen required

V = nRT/P

V = 0.02798mol × 0.082057(L atm/mol K) × 343.15K/0.9605atm

= 0.820L

Q2)

-112°C

Explanation

Volume and number of moles are constant

P1V = nRT1

P2V= nRT2

P1/P2 = T1/T2

T2 =( T1×P2) /P1

T1 = 30°C = 303.15K

P1 = 5.54atm

P2 = 2.95atm

T2 =( 303.15K × 2.95atm)/ 5.54atm = 161.4K = - 112°C

Q3)

57.93 g/mol

Explanation

PV = nRT

n /V = P/RT

n/V = molarity of the gas

P = 733mmHg = 0.9645atm

R = 0.082057(L atm/mol K)

T = 23°C = 296.15K

n/V = 0.9645atm/(0.082057 L atm/mol K × 296.15K)

n/V= 0.0397 mol/L

Density of gas = 2.30g/L

molar mass = mass/no of moles

molar mass of the gas = 2.30g/0.0397mol = 57.93g/mol