Question

3 gram H2 react with 29 gram O2 to yield H2O which is the limiting reactant calculate the maximum amount of H2O that can be formed calculate the amount of the reactant which remains unreacted

Answer 1

2H2 + O2 = 2h2O

2h2= 2*2gm=4gm
O2= 16*2= 32gm
2h20= 2*18gm

4gm of h2 react with 32g of O2
1gm of h2 react with 32/4 gram of O2
3gm of h2 react with 32/4 *3 =24gm of O2

since all three gram react with 24gm of O2,there ia no any h2 left.so it is limiting reagent.
we have 29g of o2.so unreacted(left) o2 = 29g-24g= 5g
regards.