a) The balanced reaction is : 2N2H4 (l) + N2O4 (l) ----> 4N2(g) + 8H2O(g)
Molar mass(g/mol) 32 92 28 18
b) From the above reaction 2 mol =2x32 g of N2H4 reacts with 1 mole = 92 g of N2O4
M g of N2H4 reacts with 200 g of N2O4
M = ( 2x32x200)/92 = 69.6 g of N2H4
So 100 - 69.6 = 30.4 g of N2H4 left unreacted so it is excess reactant.
Since all of N2O4 completly reacted it is limiting reactant.
Again from the reaction ,
92 g of N2O4 produces 4 mol = 4x28 g of N2
200 g of N2O4 produces ( 200x4x28)/92 = 243.5 g of N2 ---> This is the theoretical yield
c) Percentage yield = ( actual yield / theoretical yield) x 100
= ( 93.2 g / 243.5 g ) x 100
= 38.3 %
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