Chemistry help again?

Magnesium burns in air to produce magnesium oxide, MgO, and magnesium nitride, Mg3N2. Mg3N2 reacts with water to give ammonia according to the following equation:
Mg3N2 (s) + H20 (l) ----- Mg (OH) 2 (s) + NH3 (g)
1. First balance the equation
2. What volume of ammonia gas at 24 C and 753 mm Hg will be produced from 4.56 g of magnesium nitride?

Answer:
Mg3N2 +6 H2O >> 3 Mg(OH)2 + 2 NH3

Molecular weight Mg3N2 =100.9 g/mol

4.56 g / 100.9 = 0.0452 mole

The ratio between Mg3N2 and NH3 is 1 : 2 so we would get 0.0452 x2 = 0.0904 mole

T = 24 + 273 = 297 K

p= 753/760 = 0.991 atm

V = nRT/p = 0.0904 x 0.0821 x 297 / 0.991 = 2.22 L
1. Mg3N2 + 6H2O = 3Mg(OH)2 + 2NH3

2. stoicheometry of reaction

Mg3N2: 2NH3
1 : 2

# of moles in 4.56g of Mg3N2= 0.0456 mol
# of moles of NH3 released = 0.0456 * 2
=0.0912 mol

to the equation
PV=nRT
P=753 Hg mm= 753* 10^5/760 =9.9*10^4 Nm-2
T=24+273 =297 Kelvin
R=8.314 JK-1mol-1

V=0.0912mol * 8.314 JK-1 mol-1* 297 K/ 9.9*10^4 Nm-2
V=2.27*10^-3 dm-3
V=2.27 mL

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