Question

19. Nitrogen reacts with powdered aluminum according to the following chemical equation: 2 Al(s) + N2(g) → 2 AIN (S) How many liters of nitrogen gas are required to react with 22.3 g of powdered aluminum if the reaction occurs at 825 torr and 89°C?

Answer 1

The balanced equation for N₂ and aluminum is given by-

2Al (s) + N₂ (g) = 2AlN (s)

From the above equation it is very clear that 2 mol Al reacts with 1 mol N2 for complete conversion to the product.

now, 2 mol Al = 1 mol N₂ gas

moles for Al = w/M = 22.3 / 27 = 0.82 mol

[let us say molar mass for Al = 27]

2 mol Al = 1 mol N₂ gas

or, 1 mol Al = 1/2 mol N₂ gas

or, 0.82 mol of Al = 0.82 / 2 = 0.41 mol of N₂ gas

We know that PV=nRT

Here, P = 825 torr = 825 / 760 atm = 1.085 atm, temperature (T) = 89+273 = 362 K

universal gas constant (R) = 0.082 Lit atm, n = 0.41

So, volume of N₂ gas (V) = nRT/P = (0.41*0.082*362) / 1.085 = 11.21 Lit.

Therefore, 11.21 liter N₂ gas will berequired.