Question

3 g of ${{H}_{2}}$ reacts with 29 g of ${{O}_{2}}$ to yield ${{H}_{2}}O$
(i) What is limiting reagent.
(ii) Calculate the maximum amount of water that can be formed.
(iii) Calculate the amount of one of the reactants which remains unreacted.

Answer 1

Write down the reactants and products of the above reaction. Now write a balanced equation for the formation of water along with the stoichiometric coefficients. Determine the number of reactants involved for production of 1 mole of water. With this you can determine the limiting reagent as well as the maximum amount of water that can be formed. The reactant which remains unreacted will be the one in excess.

Complete answer:
(i) The reactants can be identified as hydrogen and oxygen. The product of the reaction will be water. The balanced equation is given below:
$\text{2}{{\text{H}}_{2}}\text{ + }{{\text{O}}_{2}}\text{ }\to \text{ 2}{{\text{H}}_{2}}\text{O}$
From the above reaction we can conclude that 2 moles of hydrogen are required for every mole of oxygen to form 2 moles of water.
It is given to us that 3 g of hydrogen is present with 29 g of oxygen.
Molar mass of hydrogen = 2 g
Molar mass of oxygen = 32 g
We have 1.5 moles of hydrogen and 0.9 mole of oxygen. Thus, the limiting reagent is hydrogen. This is because for 0.9 moles of oxygen, 1.8 moles of hydrogen are required.
(ii) Since 1.5 moles of hydrogen are involved in the reaction as the limiting reagent, the number of moles of water formed is 1.5 only.
(iii) 1.5 mole of hydrogen reacts with 0.75 mole of oxygen to produce water. So, the amount of oxygen left unreacted is 0.15 mole.

Note: It is important to know that the number of moles of water formed is equal to the number of moles of hydrogen reacted as long as oxygen is taken in excess. This is because the stoichiometric coefficient of hydrogen and water is the same.