Answer to Question #251941 in Inorganic Chemistry for Asri

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 → Mg₃N₂ (s)

Answer 1

Use the ideal gas equation, PV = nRT find the volume of nitrogen gas in liters from moles of N₂.

Step 1: Convert 40.0 g of Mg to moles by using the atomic mass of Mg.

Atomic mass of Mg = 24.305 g/mol

$$moles \ of \ Mg \ = \ 40.0 \cancel{\ g \ of \ Mg} \ * \ \frac{1 \ mol \ of \ Mg}{24.305 \ \cancel{g \ of \ Mg} }$$ $$= \ 1.6458 \ mol \ of \ Mg.$$

Step 2: Using the mol to mol ratio of Mg and N₂ from the balanced chemical reaction find the moles of N₂ that will react with 1.6458 moles of N₂.

  3Mg (s) + N₂ → Mg₃N₂ (s)

The mol to mol ratio of Mg and N₂ in this balanced chemical reaction is 3: 1.

$$moles \ of \ Mg \ = \ 1.6458 \cancel{\ mol \ of \ Mg} \ * \ \frac{1 \ mol \ of \ N_2}{3 \ \cancel{mol \ of \ Mg} }$$ $$= \ 0.5486 \ mol \ of \ N_2$$

Step 3: Using the ideal gas equation, PV = nRT find the volume of nitrogen gas in liters occupied by 0.5486 moles of N₂ at a pressure of 2.25 atm and a temperature of 15 °C.

Pressure, P = 2.25 atm,

Temperature, T in Kelvin = 15 °C = 15 + 273.15 = 288.15 K

moles of N₂ , n = 0.5486 moles of N₂

Gas constant, R = 0.08206 L.atm/mol.K

plug all this information in the ideal gas equation we have

$2.25 \ atm * V \ = \ 0.5486 \ mol \ * \ 0.08206 \ L.atm/mol.K * 288.15 \ K$

$2.25 \ atm * V \ = \ 12.9716 \ L.atm$

dividing both the side by 2.25 atm, we have

$$\frac{\cancel{2.25 \ atm} * V}{ \cancel{2.25 \ atm}} = \ \frac{12.9716 \ L.atm}{2.25 \ atm }$$

$$V = 5.765 \ L$$