10g of hydrogen and 64g of oxygen were filled in a steel vessel and exploded. Amount of water produced in this reaction will be:-
(a) 3mol
(b) 4mol
(c) 1mol
(d) 2 mol

Hint: This question is based on the concept of limiting reagent and excess reagent. For calculating the moles given mass should be divided with molecular mass. This can be done with the help of a properly balanced chemical equation of the reaction.

Complete step by step answer:
Let us first understand about limiting reagent and excess reagent.
In some chemical reactions, one of the reactants is present in a larger amount than the other as required. Then the amount of the product formed depends upon the reactant which has reacted completely.
The reactant which has completely used and whose amount is less is called the limiting reagent or the limiting reactant.
The reactant which is not used completely and whose amount is left after the reaction is called excess reagent or excess reactant.
So, the question says hydrogen reacts with oxygen to form water, the chemically balanced equation will be:
\[\begin{align}
  & 2{{H}_{2}}\text{ }+\text{ }{{O}_{2}}\to \text{ }2{{H}_{2}}O \\
 & 2moles\text{ 1mole 2moles} \\
\end{align}\]
For calculating the number of moles: given mass must be divided with molecular mass.
For hydrogen, the molecular mass is 2.
10g of hydrogen reacts, the moles = \[\dfrac{10}{2}=5moles\]
For oxygen, the molecular mass is 32.
64g of oxygen reacts, the moles = \[\dfrac{64}{32}=2moles\]
According to the balanced equation, 2 moles of hydrogen reacts with 1 mole of oxygen to form 2 moles of water as a product.
Hence, the oxygen acts as the limiting reactant and the hydrogen acts as the excess reactant.
2 moles of oxygen are present in the reaction, it will combine with 2*2 = 4 moles of hydrogen.
It will produce 4 moles of water.
Hence, option (b)- 4 moles is correct.

Note: The limiting and excess reactant can only be calculated with the number of moles. The chemical equation must be balanced properly otherwise you would get the wrong result.