Given the following amounts of gases calculate the number of moles of each gas. Calculate the volume each amount of gas would occupy at 100°C and 15 atm. a. 5.8g NH₃ b. 48g O₂

Md Ariful Islam

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Solution:

The molar mass of NH3 is 17.03 g/mol

The molar mass of O2 is 32.00 g/mol

Calculate the moles of each gas:

a) (5.8 g NH3) × (1 mol NH3 / 17.03 g NH3) = 0.34 mol NH3

Moles of NH3 = 0.34 mol

b) (48 g O2) × (1 mol O2 / 32.00 g O2) = 1.50 mol O2

Moles of O2 = 1.50 mol

The Ideal Gas equation can be used: PV = nRT

To find the volume each amount of gas, solve the equation for V:

V = nRT / P

R = 0.0821 L atm mol-1 K-1

T = 100°C = 373.15 K

a) V(NH3) = (0.34 mol × 0.0821 L atm mol-1 K-1 × 373.15 K) / (15 atm) = 0.6944 L = 0.69 L

Volume of NH3 = 0.69 L

b) V(O2) = (1.50 mol × 0.0821 L atm mol-1 K-1 × 373.15 K) / (15 atm) = 3.0636 L = 3.06 L

Volume of O2 = 3.06 L

Answers:

a) 0.34 mol NH3; 0.69 L NH3

b) 1.50 mol O2; 3.06 L O2

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