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# If 30ml of ${{H}_{2}}$and 20ml of ${{O}_{2}}$ reacts to form ${{H}_{2}}O$, what is left at the end of the reaction?This question has multiple correct options:A. 10ml of ${{H}_{2}}$B. 5ml of ${{H}_{2}}$C. 10ml of ${{O}_{2}}$D. 5ml of ${{O}_{2}}$

Last updated date: 11th Jun 2024
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Hint: The reactant that fully reacts in the reaction is called reactant limiting or reagent limiting. The reactant that is not fully consumed in the reaction is called excess reactant.You should write the balanced reaction equation before solving the question. A balanced equation is an equation for a chemical reaction in which the number of atoms in the reaction for each element and the total charge for both the reactants and the products is equal.

The Law of Definite Proportions, states that any chemical compound will always contain a fixed ratio of elements by mass. The Law of Definite Proportions is also sometimes called Proust's Law.
${{H}_{2}}$and ${{O}_{2}}$react to form water.
The reaction used in question is: $2{{H}_{2}}(g)+{{O}_{2}}(g)\xrightarrow{{}}2{{H}_{2}}O(g)$
2 moles of ${{H}_{2}}$combines with 1 mole of ${{O}_{2}}$ to produce 2 moles of ${{H}_{2}}O$.
In other words, we can say that 1ml of ${{H}_{2}}$reacts with 0.5ml of ${{O}_{2}}$ to form 1ml of ${{H}_{2}}O$.
Hence, when 30ml of ${{H}_{2}}$reacts with 15ml of ${{O}_{2}}$to form water it can be represented in a balanced chemical reaction as:
${{H}_{2}}(g)+\dfrac{1}{2}{{O}_{2}}(g)\xrightarrow{{}}{{H}_{2}}O(g)$

Therefore, limiting reagent is ${{H}_{2}}$.
So, at the end of the reaction, 30ml of water and 5ml of ${{O}_{2}}$will be left.
So, the correct answer is “Option D”.

Note: The limiting reagent (also known as limiting reactant) in a chemical reaction is a reactant that is totally consumed when the chemical reaction is completed. The amount of product formed is limited by this reagent, since the reaction cannot continue without it.
Remember this reaction equation,
$2{{H}_{2}}+{{O}_{2}}\xrightarrow{{}}2{{H}_{2}}O$
When 2 moles of ${{H}_{2}}$combines with 1 mole of ${{O}_{2}}$ to produce 2 moles of ${{H}_{2}}O$.