(D) The balanced chemical equation for the reaction is: $2H_2(g) + O_2(g) \to 2H_2O(l)$.According to the stoichiometry,$2 \ mL$ of $H_2$ reacts with $

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(D) The balanced chemical equation for the reaction is: $2H_2(g) + O_2(g) \to 2H_2O(l)$.According to the stoichiometry,$2 \ mL$ of $H_2$ reacts with $1 \ mL$ of $O_2$.Therefore,$30 \ mL$ of $H_2$ will react with $\frac{1}{2} \times 30 \ mL = 15 \ mL$ of $O_2$.Since we started with $20 \ mL$ of $O_2$,the remaining amount of $O_2$ is $20 \ mL - 15 \ mL = 5 \ mL$.Thus,$5 \ mL$ of $O_2$ is left at the end of the reaction.

Class 11 Chemistry Some Basic Concepts of Chemistry Chemical equation and limiting reagent

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If $30 \ mL$ of $H_2$ and $20 \ mL$ of $O_2$ react to form water,what is left at the end of the reaction?

A
$10 \ mL$ of $H_2$
B
$5 \ mL$ of $H_2$
C
$10 \ mL$ of $O_2$
D
$5 \ mL$ of $O_2$

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Some Basic Concepts of Chemistry

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Chemical equation and limiting reagent

When $10 \, mL$ of $10 \, M \, H_2SO_4$ solution is mixed with $100 \, mL$ of $1 \, M \, NaOH$ solution,the resulting solution will be .......

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The number of moles of $H_2$ required to produce $10$ moles of $NH_3$ in the following reaction is:
$a H_{2(g)} + b NO_{2(g)} \longrightarrow c NH_{3(g)} + d H_2O_{(g)}$

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In the preparation of baking soda,what is the ratio of $H_2O$ and $CO_2$ used to react with $Na_2CO_3$ to form $NaHCO_3$?

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The mass of $N_2F_4$ produced by the reaction of $2.0 \ mol$ of $NH_3$ and $8.0 \ mol$ of $F_2$ is $0.5 \ mol.$ What is the per cent yield?
$2NH_3 + 5F_2 \to N_2F_4 + 6HF$

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If the heat of combustion of methane is $-100 \ kJ/mol$,then calculate the heat evolved when $32 \ g$ of methane reacts with $32 \ g$ of oxygen. (in $kJ$)

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If $30 \ mL$ of $H_2$ and $20 \ mL$ of $O_2$ react to form water,what is left at the end of the reaction?

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When $10 \, mL$ of $10 \, M \, H_2SO_4$ solution is mixed with $100 \, mL$ of $1 \, M \, NaOH$ solution,the resulting solution will be .......

The number of moles of $H_2$ required to produce $10$ moles of $NH_3$ in the following reaction is:$a H_{2(g)} + b NO_{2(g)} \longrightarrow c NH_{3(g)} + d H_2O_{(g)}$

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The number of moles of $H_2$ required to produce $10$ moles of $NH_3$ in the following reaction is:
$a H_{2(g)} + b NO_{2(g)} \longrightarrow c NH_{3(g)} + d H_2O_{(g)}$

The mass of $N_2F_4$ produced by the reaction of $2.0 \ mol$ of $NH_3$ and $8.0 \ mol$ of $F_2$ is $0.5 \ mol.$ What is the per cent yield?
$2NH_3 + 5F_2 \to N_2F_4 + 6HF$