Question

The combustion of ammonia in the presence of excess oxygen yields NO_2 and H_2O: 4 NH_3(g)+7O_2(g) right arrow 4 NO_2(g)+6H_2O(g) The combustion of 43.9 g of ammonia produces g of NO_2.

Answer 1

4NH3(g) + 7O2(g) ------ 4NO2(g) + 6H2O(g)

Molar mass of NH3 = 14 + 3 * 1 = 17 gm/mol

Number of moles of ammonia = 43.9/17 = 2.582 moles

4 moles of NH3 will give 4 moles of NO2

Hence moles of NO2 formed = 2.582 moles

Molar mass of NO2 = 14 + 2 * 16 = 46 gm/mol

Mass of NO2 formed = 2.582 moles * 46 gm/mol

=> 118.788 gms