Question

If 50.0 g of oxygen and 50.0 g of hydrogen react to produce water.

a. Write a balanced equation for the reaction.

b. How many moles of oxygen are in 50.0 g?


c. How many moles of water could be produced from 50.0 g of oxygen and excess hydrogen?

d. How many moles of hydrogen are in 50.0 g ?!

Answer 1

m(O2)=50g

m(H2)=50g

m(H2O)-?

n2(H2)-?

2H2 + O2 = 2H2O

n(O2)= m (O2)/M(O2) =50g / 32 g/mol= 1,56 mol.

n(H2)= m (H2)/M(H2) =50g / 2 g/mol= 25 mol.

Since oxygen gas is the limiting reactant,

n(O2)< 2 n(H2) from reaction.

n(H2O)= 2n(O2)= 2n (H2 reac.)=2*1,56 mol=3,12mol.

m(H2O)=n(H2O)*M(H2O)= 3,12mol* 18 g/mol.

n2(H2)= n(H2) - n (H2 reac.)=25mol - 3,12mol=21,88mol.