Question

The volume of carbon dioxide gas evolved at $S.T.P.$ by heating $7.3g$ of $Mg{\left( {HC{O_3}} \right)_2}$ will be:
A: $2240ml$
B: $1120ml$
C: $2340ml$
D: $2000ml$

Answer 1

$S.T.P.$ refers to standard temperature and pressure. At this condition one mole of gas occupies volume of $22400ml$. To find the volume of carbon dioxide gas evolved in this question we have to write the reaction taking place and then after comparing the number of moles we can find the volume of carbon dioxide gas.
Formula used: Number of moles$= \dfrac{{{\text{given mass}}}}{{{\text{molecular mass}}}}$

Formula used:

Complete answer: or Complete step by step answer:
In this question we have to find the volume of carbon dioxide gas evolved when $7.3g$ of $Mg{\left( {HC{O_3}} \right)_2}$ will be heated. Reaction taking place when $Mg{\left( {HC{O_3}} \right)_2}$ is heated is:
$Mg{\left( {HC{O_3}} \right)_2}\xrightarrow[{}]{}MgO + {H_2}O + 2C{O_2}$
From the above reaction we can see that when one mole of $Mg{\left( {HC{O_3}} \right)_2}$ is heated, two moles of carbon dioxide gas is released.
Molecular mass is defined as the mass of all the atoms present in a molecule. Molecular mass of $Mg{\left( {HC{O_3}} \right)_2}$ is $146$ (atomic mass of magnesium is $24$, hydrogen is $1$, carbon is $12$ and oxygen is $16$.
Molecular mass of carbon dioxide $\left( {C{O_2}} \right)$ is $44$. From the reaction we can write,
$146g$ of $Mg{\left( {HC{O_3}} \right)_2} = 2 \times 44g$ of $C{O_2}$
$1g$ of $Mg{\left( {HC{O_3}} \right)_2} = \dfrac{{88}}{{146}}g$ of $C{O_2}$
Given mass of $Mg{\left( {HC{O_3}} \right)_2}$ is $7.3g$. Therefore,
$7.3g$ of $Mg{\left( {HC{O_3}} \right)_2} = \dfrac{{88}}{{146}} \times 7.3g$ of $C{O_2}$
Solving this we get mass of $C{O_2}$ equal to $4.4g$.
Molecular mass of carbon dioxide $\left( {C{O_2}} \right)$ is $44$. Therefore number of moles of carbon dioxide $\left( {C{O_2}} \right)$ formed will be $\dfrac{{4.4}}{{44}} = 0.1$ (using the formula Number of moles$= \dfrac{{{\text{given mass}}}}{{{\text{molecular mass}}}}$)
We know that one mole of gas at $S.T.P.$ occupies $22400ml$. It can also be written as:
One mole$= 22400ml$
Moles obtained are $0.1$.
$0.1$ moles$= 22400 \times 0.1 = 2240ml$

So, the correct answer is Option A .

Note:
At $S.T.P.$ temperature of a gas is equal to $273.25K$ and pressure of the gas is equal to $1atm$. One mole of a gas at $S.T.P.$ occupies $22400ml$. In one mole of a substance there are $6.022 \times {10^{23}}$ particles. Number of particles in one moles is also called Avogadro number.