# Answer to Question #227763 in Inorganic Chemistry for john

Question #227763

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)

1
2021-08-20T01:51:46-0400

Q227763

By referring to the chemical equation given, calculate the volume of nitrogen gas in liters, 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)

Solution :

First, convert 40.0 g of magnesium to moles by using the atomic mass of Mg.

Next using the mol to mol ratio of Mg and N2 from the balanced chemical reaction find the moles of N2 that will react with calculated moles of N2.

Lastly using 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

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.

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

dividing both the side by 2.25 atm, we have

In the correct significant figure, the answer is 5.77 L.

So we will require 5.77 L of N2 to react completely with 40.0 g of magnesium at a pressure of 2.25 atm and a temperature of 15 °C.

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