Question

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2.0g of CH4 (16.0) are added to 3.0g of O2 (32.0) at 400K and 950torr. What volume does this mixture occupy?

A 300 ml sample of gas was collected at STP weighs .90g. What is the molar mass of this gas?​

Answer 1

Answer:

1) 5.74 L.

2) 67.158 g/mol.

Explanation:

Q1:

• We suppose that the two gases CH₄ and O₂ behave ideally.

• We can use the general gas law of ideal gases:

PV = nRT

P is the pressure of the gases (P = 950 torr / 760 = 1.25 atm).

V is the volume of the gases (V = ??? L).

R is the general gas constant (R = 0.082 L.atm/mol.K).

T is the temperature of the gases (T = 400 K).

n is the no. of moles of the gases.

• n = no. of moles of CH₄ + no. of moles of O₂.

no. of moles of CH₄ = mass / molar mass = (2.0 g) / (16.0 g/mol) = 0.125 mol.

no. of moles of O₂ = mass / molar mass = (3.0 g) / (32.0 g/mol) = 0.09375 mol.

∴ n = no. of moles of CH₄ + no. of moles of O₂ = 0.125 mol + 0.09375 mol =  0.21875 mol.

∴ V = nRT/P = (0.21875 mol)(0.082 L.atm/mol.K)(400 K) / (1.25 atm) = 5.74 L.

Q2:

• We can also use the general gas law: PV = nRT.

• Each term in the law is defined in the answer of the first question.

• n = mass / molar mass, we substitute by this value in the gas law.

PV = (mass/molar mass)RT.

The molar mass = (mass)RT/PV.

The mass of the gas = 0.90 g.

R = 0.082 L.atm/mol.K.

T = 273 K at STP conditions.

P = 1.0 atm at STP conditions.

V = 300.0 mL = 0.30 L.

• The molar mass = (mass)RT/PV = (0.90 g)(0.082 L.atm/mol.K)(273.0 K) / (1.0 atm)(0.30 L) = 67.158 g/mol.