Question

Calcium carbonate reacts with aqueous HCl to give $CaCl_{2}$ and $CO_{2}$ according to the reaction:
$CaCO_{3}(s)+2HCL(aq)$ \longrightarrow $CaCl_{2}(aq) +CO_{2} (g)+H_{2}O(l)$
What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl?

Answer 1

$Ca{ CO }_{ 3(s) }\quad +\quad 2H{ Cl }_{ (aq) }\quad \rightarrow \quad { CaCl }_{ 2(aq) }\quad +\quad { CO }_{ 2(g) }\quad +\quad { H }_{ 2 }{ O }_{ (g) }$

i) Calculating the mass of HCl in 25 ml of 0.75M HCl

We know that,

$Molarity\quad =\quad \frac { Mass\quad of\quad solute }{ Volume\quad of\quad solution\quad in\quad L }$

$0.75\quad =\quad \frac { Mass\quad of\quad HCl\quad \times \quad 1000\quad mL }{ 25\quad mL }$

Mass of HCl = 0.6844 g  

ii) Calculating of mass of $Ca{ CO }_{ 3 }$

From the above equation we know that,  

2 mole of HCl reacts with 1 mole of $Ca{ CO }_{ 3 }$.

$2\quad \times \quad 36.5 g$ of HCl reacts with 100 g of $Ca{ CO }_{ 3 }$.

73 g of HCl reacts with 100 g of $Ca{ CO }_{ 3 }$.

1 g of HCl reacts with  of $Ca{ CO }_{ 3 }$.

0.6844 g of HCl reacts with $100\quad \times \quad \frac { 0.6844 }{ 73 } \quad g\quad of\quad Ca{ CO }_{ 3 }\quad =\quad \frac { 68.44 }{ 73 } \quad g\quad =\quad 0.9375\quad g$

0.9375 g of $Ca{ CO }_{ 3 }$ is required to react completely with 25ml of 0.75M HCl.