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)
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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
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