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Magnesium reacts with N2 and O2 to form MgO Mg3N2. calculate the mass of nitrides and...

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.

Solutions

Expert Solution

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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