Answer to Question #122132 in General Chemistry for Emilie

Question

If 10.3 g of magnesium nitride is treated with 10.3 g of water, what volume of ammonia gas would be collected at 24˚C and 0.989 atm?

Answer 1

Mg₃N₂ + 6H₂O --> 3Mg(OH)₂ + 2NH₃

n (Mg₃N₂) = 10.3g / 100.9g/mol = 0.102 mol

n (H₂O) = 10.3g / 18.0g/mol = 0.572 mol

According to the equation, the Mg₃N₂ : H₂O mole ratio is 1 : 6

The actual ratio is 1 : (0.572/0.102) = 1 : 5.6. Therefore H₂O is the limiting reactant.

According to the equation, n(NH₃) = n(H₂O) / 3 = 0.191 mol

pV = nRT ==> V = nRT / p

T = 24^oC = 297K

p = 0.989 atm = 100.2 kPa

V = (0.191mol x 8.31J/mol*K x 297K) / 100.2kPa = 4.7 L

Answer: 4.7 L