(D) The balanced chemical equation is: $Ca + 2H_2O \to Ca(OH)_2 + H_2$From the stoichiometry,$1 \ mol$ of $Ca$ $(40 \ g)$ produces $1 \ mol$ of $H_2$

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(D) The balanced chemical equation is: $Ca + 2H_2O \to Ca(OH)_2 + H_2$From the stoichiometry,$1 \ mol$ of $Ca$ $(40 \ g)$ produces $1 \ mol$ of $H_2$ gas.$1 \ mol$ of any gas at $STP$ occupies $22400 \ cm^3$.Given mass of $Ca = 8 \ g$.Moles of $Ca = \frac{8 \ g}{40 \ g/mol} = 0.2 \ mol$.Since $1 \ mol$ of $Ca$ produces $1 \ mol$ of $H_2$,$0.2 \ mol$ of $Ca$ will produce $0.2 \ mol$ of $H_2$.Volume of $H_2$ at $STP = 0.2 \ mol \times 22400 \ cm^3/mol = 4480 \ cm^3$.

Class 11 Chemistry Some Basic Concepts of Chemistry Chemical stoichiometry

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The reaction of calcium with water is represented by the equation $Ca + 2H_2O \to Ca(OH)_2 + H_2$. What volume of $H_2$ at $STP$ would be liberated when $8 \ g$ of calcium completely reacts with water? .......... $cm^3$

AIIMS 2010 All AIIMS

A
$0.2$
B
$0.4$
C
$2240$
D
$4480$

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Some Basic Concepts of Chemistry

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Chemical stoichiometry

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The reaction of calcium with water is represented by the equation $Ca + 2H_2O \to Ca(OH)_2 + H_2$. What volume of $H_2$ at $STP$ would be liberated when $8 \ g$ of calcium completely reacts with water? .......... $cm^3$

Similar Questions

$A$ cylinder of gas is assumed to contain $11.2 \ kg$ of butane $(C_4H_{10})$. If a normal family needs $20000 \ kJ$ of energy per day,the cylinder will last: (Given that $\Delta H$ for combustion of butane is $-2658 \ kJ/mol$)....$days$

Combustion of glucose $(C_6H_{12}O_6)$ produces $CO_2$ and water. The amount of oxygen (in $g$) required for the complete combustion of $900 \ g$ of glucose is:[Molar mass of glucose in $g \ mol^{-1} = 180$]

Calculate the volume of $H_2$ gas kept at $STP$ if it contains as many $H$ atoms as in $98 \ g$ of $H_3PO_4$. (Atomic mass of $P = 31 \ u$) (in $L$)

The volume of oxygen at $STP$ required to burn $2.4 \,g$ of carbon completely is $.... \,L$

The molarity of a $NaOH$ solution prepared by dissolving $4 \ g$ of it in $250 \ mL$ of water is: (in $M$)

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Combustion of glucose $(C_6H_{12}O_6)$ produces $CO_2$ and water. The amount of oxygen (in $g$) required for the complete combustion of $900 \ g$ of glucose is:
[Molar mass of glucose in $g \ mol^{-1} = 180$]