Question

By referring to the chemical equation given, calculate the volume of nitrogen gas in litres, required to react with 40.0 g of magnesium at a pressure of 2.25 atm and a temperature of 15 °C.

3Mg (s) + N2 à ƒ ƒ ¢ † ’ Mg3N2 (s)

Answer 1

3Mg + N₂ → Mg₃N₂

M(Mg) = 24.3 g/mol

n(Mg) $= \frac{40.0}{24.3} = 1.646 \;mol$

According to the reaction equation

n(N₂) $= \frac{1}{3}n(Mg) = \frac{1}{3} \times 1.646 = 0.548 \;mol$

Ideal gas low

pV = nRT

p= 2.25 atm

T=15 + 273 = 288 \;K

R = 0.08206 L×atm/mol×K

$V = \frac{nRT}{p} \\ V (N_2) = \frac{0.548 \times 0.08206 \times 288}{2.25} = 5.756 \;L$

Answer: 5.756 L