The Balanced reaction 3H2(g) + N2(g) = 2NH3(g) was carried out in 57.1% yield, with 10.4g of NH3 collected as the product. If hydrogen, H2 was the limiting reactant, how many moles of H2, were added at the beginning of the reaction?

Please show all calc involved

Question

The Balanced reaction 3H2(g) + N2(g) = 2NH3(g) was carried out in 57.1% yield, with 10.4g of NH3 collected as the product. If hydrogen, H2 was the limiting reactant, how many moles of H2, were added at the beginning of the reaction?

Please show all calc involved

anonymous · Answer

Moles of NH3=10.4/17=0.611
for 100precent yield 1mole of H2 gives 2/3 moles of NH3
for 57.1% yild 1mole of H2 will give 0.571*2/3=0.38
1 mole gives 0.38
number of moles for 0.61=0.61/0.38=1.605