Question

Calcium carbonate reacts with aqueous $HCl$ to gives $CaC{l_2}$ and $C{O_2}$ according to the reaction $CaC{O_3}\left( s \right) + 2HCl\left( {aq} \right) \to CaC{l_2}\left( {aq} \right) + C{O_2}\left( g \right) + {H_2}O\left( l \right)$ . The mass of $CaC{O_3}$ which is required to react completely with $25ml$ of $0.75M$$HCl$ is:
$\left( A \right)$ $1.825g$
$\left( B \right)$ $0.9375g$
$\left( C \right)$ $1.8357g$
$\left( D \right)$ $0.46875g$

Answer 1

Hint: First we have to know the molar mass of a chemical compound is defined as the mass of a sample of that compound divided by the amount of substance in that sample, measured in moles. The molar mass is a bulk, not molecular, property of a substance. The molar mass of a compound is calculated by adding there standard atomic masses in \[\left( {\dfrac{g}{{mol}}} \right)\] of each atom.

Complete step by step answer:
Since, we have to calculate the mass of calcium carbonate $\left( {CaC{O_3}} \right)$ , Let us calculate the mass of calcium carbonate $\left( {CaC{O_3}} \right)$ is $x$ grams.
Molar mass $x$ of $CaC{O_3}$ is the sum of the masses of calcium $ + $carbon $ + $$3$ oxygen $ = $ $40 + 12 + 48$$ = 100g$ .
In the calcium carbonate$\left( {CaC{O_3}} \right)$, the molar mass is $100g$
$1000ml$ Of $0.75HCl$ $ = 0.75$ mole
So, \[25ml\] of hydrochloric acid $\left( {HCl} \right)$ will contain \[HCl\]$ = 0.75 \times \dfrac{{25}}{{1000}} = 0.01875$
$2$ mol of hydrochloric acid $\left( {HCl} \right)$reacts with $1$ mol of $CaC{O_3}$
So, $0.01875$ mol of hydrochloric acid $\left( {HCl} \right)$ will react with $\dfrac{1}{2} \times 0.01875 = 0.009375$
Molar mass of calcium carbonate $\left( {CaC{O_3}} \right)$ $ = 100g$
Hence, the mass of $0.009375$ mol of calcium carbonate $CaC{O_3} = $ no.of moles $ \times $ molar mass
Mass of calcium carbonate $\left( {CaC{O_3}} \right)$ $ = 0.009375 \times 100$ g .
$ = 0.9375g$ .
Hence, the correct option is $\left( C \right)$ .

Additional information: The molar mass is the mass of a given chemical element or chemical compound $\left( g \right)$ divided by the amount of substance $\left( {mol} \right)$ .

Note:
We must know the definition of atomic mass unit is the atomic mass of an element is the average mass of the atoms of an element measured in atomic mass unit (amu, also as Daltons, D). Where $1\,\,amu = 1.6605 \times {10^{ - 24}}g$ .