Question

56 kg of N2(g) and 10 kg of H2(g) are mixed to produce NH3(g). Calculate the number of moles of

ammonia gas formed. (atomic mass of N = 14, H = 1)​

Answer 1

Answer:

Explanation:

By equation

N2+3H2--------2NH3

it's a limiting reagent question,let's first find limiting and excess reagent

1mole of N2 reacts with 3 miles of H2 to form ammonia

Moles in 56 KG N2-------56000÷28 (14molar mass ×2)=2000moles

Moles in 10kg of hydrogen=10000÷2(2 ×1molecules of hydrogen)=50000moles

1moles reacts ------3 moles of H2

2000moles --------2000×3=6000moles

But given moles of H2 are 50000 ,therefore hydrogen here is excess reagent and nitrogen is limiting reagent

1moles of N2 ------forms 2 moles of NH3

2000moles of N2 will form --------2×2000=4000moles of NH3

Mass of NH3=4000×molar mass of NH3(AS given mass ÷molar mass =moles )=4000×17(14plus 3)=68000grams /68kg of ammonia