Question

Calcium carbonate reacts with aqueous HCI to give $CaC{l}_2$ according to the reaction,

$CaC{O}_3\left(s\right)+2HCl\left(aq\right)-Cac{l}_2\left(aq\right)+C{O}_2\left(g\right)+H_{2}O(l.)$

What mass of $CaC{O}_3$ is required to react completely with $25mL\text{of}0.75MHCl?$

Answer 1

Calculation of amoun of $HCl$

$0.75MHCl\text{means}0.75\text{mol of}HCl\text{are present in}1L\text{of water.}$

We know that, molar mass of $HCl=36.5gmo{l}^{-1}$

So, mass of $HCl$ present in $1L$ of water

$=\left(0.75mol\right)\times (36.5gmo{l}^{-1}=27.375g.)$

$Thus,1000mL\text{of solution contains}27.375g\text{of HCl.}$

Therefore, mass of $HCl$ present in $25mL$ of solution.

$=\frac{27.375g}{1000mL}\times 25mL$

$=0.6844g$

Calculation of mass of $CaC{O}_3$ required.
Given balanced chemical equation is:

$CaC{O}_3\left(s\right)+2HCl\left(aq\right)\to CaC{l}_2\left(aq\right)+C{O}_2\left(g\right)H_{2}O\left(l\right)$

So,$2\text{moles of HCl reacts with}1\text{mole of}CaC{O}_3$

Hence,$73g\text{of HCl reacts with}100g\text{of}CaC{O}_3$

Therefore, amount of $CaC{O}_3$ that is required to react with

$0.6844g\text{of}HCl$

$=\frac{100}{73}\times 0.6844g$
$=0.9375g$

Final answer: Amount of $CaC{O}_3$ that will react with $0.6844gHCl$ according to balenced chemical equation $=0.9375g.$