Question

Nitrogen reacts with powdered aluminum according to the reaction:
2Al(s)+N2(g)→2AlN(s)

How many liters of N2 gas, measured at 896 torrtorr and 96 ∘C∘, are required to completely react with 18.6 g of Al?

Answer 1

2Al(s) + N2(g) >>>> 2AlN(s)

2 mol Al(27g/mol) reacts with 1mol N2(28g/mol) gas completely.

Means , 54g Al reacts completely with 28g N2 .

Then , for 18.6g Al we need N2 ;

(28g-N2/54g-Al) × 18.6g-Al = 9.64g N2 is required.

Moles of N2 required = mass/molar mass = 9.64g/28g/mol = 0.344mol of N2 .

As per ideal gas equation;

PV = nRT

Where, P = pressure of gas in atm , 896torr × (1atm/760torr) = 1.18atm

V = liter volume of gas

n = moles of gas , 0.344mol

R = gas constant , 0.0821Latm/mol.K

T = temperature in Kelvin, 96°C = 273.15+96 = 369.15K

Using all values to calculate (V) volume of gas;

V = nRT/P = (0.344×0.0821×369.15)÷1.18 = 8.84L

Thus we require 8.84L of N2 gas for complete reaction of 18.6g of Al .