Answer to Question #252548 in Inorganic Chemistry for Sanchanaa

Question #252548
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)
1
Expert's answer
2021-10-17T11:19:03-0400

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} \\\\\n\nV (N_2) = \\frac{0.548 \\times 0.08206 \\times 288}{2.25} = 5.756 \\;L"

Answer: 5.756 L


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