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)

  

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Expert's answer
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



moles of Mg = 40.0 g of Mg  1 mol of Mg24.305 g of Mgmoles \ of \ Mg \ = \ 40.0 \cancel{\ g \ of \ Mg} \ * \ \frac{1 \ mol \ of \ Mg}{24.305 \ \cancel{g \ of \ Mg} }


= 1.6458 mol 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 mol of Mg  1 mol of N23 mol of Mgmoles \ of \ Mg \ = \ 1.6458 \cancel{\ mol \ of \ Mg} \ * \ \frac{1 \ mol \ of \ N_2}{3 \ \cancel{mol \ of \ Mg} }


= 0.5486 mol of N2= \ 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 atmV = 0.5486 mol  0.08206 L.atm/mol.K288.15 K2.25 \ atm * V \ = \ 0.5486 \ mol \ * \ 0.08206 \ L.atm/mol.K * 288.15 \ K


2.25 atmV = 12.9716 L.atm2.25 \ atm * V \ = \ 12.9716 \ L.atm


dividing both the side by 2.25 atm, we have



2.25 atmV2.25 atm= 12.9716 L.atm2.25 atm\frac{\cancel{2.25 \ atm} * V}{ \cancel{2.25 \ atm}} = \ \frac{12.9716 \ L.atm}{2.25 \ atm }



V=5.765 LV = 5.765 \ L

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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