Question

Magnesium reacts with N2 and O2 to form MgO Mg3N2. calculate the mass of nitrides and oxides, which are obtainable by reacting one liter of air at 296 K and 749 mmHg, and an excess of magnesium.

Answer 1

3Mg+N₂--------> Mg₃N₂

Mg+0.5O₂ -----------> MgO

Calculate the moles of air:

pressure in atm 0.9855/740 mmHg

n=PV/RT=(0.9855 atm*1 L)/(0.0821 atm L/mol K)*296 K.=0.04055 mol

1 mol of air contains 21 % oxygen and 79 % nitrogen.

Moles of oxygen =0.21*0.04055=0.0085 mol

Moles of nitrogen=0.79*0.04055=0.0320 mol

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1 mol of Nitrogen requires 3 moles of Mg

Moles of magnesium=0.032*3=0.096 mol, N2 is the limiting reactant.

Moles of Mg₃N₂ , 1 mol of N2 produces 1 mol of Mg₃N₂

Moles of Mg₃N₂ =0.032 mol,molar mass of Mg₃N₂ is100.9494 g/mol

Mass of  Mg₃N₂ =3.2 g

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1 mol of Mg requires 0.5 mol of Oxygen

Moles of magnesium =0.0085/0.5=0.017 mol.

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O2 is the limiting reactant.

0.5 mol of O2 produces 1 mol of MgO

0.0085 mol mol of O2

Mgo moles = 0.017 mol

Molar mass: 40.3044 g/mol

Mass of MgO=0.6851 g

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