Question

How many moles of N2 can form when 0.412 mol of N2H4(l) reacts with 0.195 mol of N2O4(l) in the following reaction 2N2H4(l) + N2O4(l) → 3N2(g) + 4H2O(g) when the yield is 78.2%?

Answer 1

2N2H4(l) + N2O4(l) → 3N2(g) + 4H2O(g)
according to reaction
1 mol of N2O4 required 2 mol of N2H4
0.195 mol of N2O4 required 0.390 mol of N2H4
but we have 0.412 mole of N2H4
so, N2H4 is in excess
and N2O4 is limiting reagent

again,
1 mol of N2O4 give 3 mol of N2
0.195 mol of N2O4 give (3*0.195) mol of N2
number of mole of N2 formed = 0.585 mole

yield = {(actual mole)/(theoretical mole)}*100
78.2 = {(actual mole)/0.585}*100
(actual mole) = 0.458 mole

Answer : 0.458 mole