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
Answer
578.9k+ views
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.
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.
Recently Updated Pages
Disproportionation Reaction: Definition, Example & JEE Guide

JEE Main 2025-26 Mock Test: Binomial Theorem & Quick Practice

Properties of Solids and Liquids Mock Test 2025

JEE Main Mock Test 2025-26: Principles Related To Practical

JEE Main Mock Test 2025-26: Dual Nature of Matter & Radiation

JEE Main 2025-26 Work, Energy and Power Mock Test – Free Practice Online

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Understanding the Electric Field of a Uniformly Charged Ring

Understanding Atomic Structure for Beginners

Understanding the Different Types of Solutions in Chemistry

Derivation of Equation of Trajectory Explained for Students

Electron Gain Enthalpy and Electron Affinity Explained

Other Pages
JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

JEE Advanced 2026 Notification Out with Exam Date, Registration (Extended), Syllabus and More

NCERT Solutions For Class 11 Chemistry In Hindi Chapter 1 Some Basic Concepts Of Chemistry - 2026-27

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding Instantaneous Velocity

Understanding Electromagnetic Waves and Their Importance

