CaCO 3 is decomposed by HCl (density 1.825 g / cc ) CaCO 3+2 HCl arrow CaCl 2+ H 2 O + CO 2 Volume of HCl required to decompose 10 g of 50 % pure CaCO 3 is

Question

CaCO 3 is decomposed by HCl (density 1.825 g / cc ) CaCO 3+2 HCl arrow CaCl 2+ H 2 O + CO 2 Volume of HCl required to decompose 10 g of 50 % pure CaCO 3 is (A) 1.825 mL (B) 3.65 mL (C) 0.9125 mL (D) 2 mL

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/fb302d302b9c547c676f636c0599acbc-.png" /> (50 % pure) 5 g (7.3/2) g HCl ( text mass / text volume )= density ∴ Volume of HCl =( text mass of HCl / text density of HCl )=(7.3/2 × 1.825) =2 mL

Q. $C a C O_{3}$ is decomposed by $H Cl$ (density $1.825 g / cc$ )
$C a C O_{3} + 2 H Cl \to C a C l_{2} + H_{2} O + C O_{2}$
Volume of $H Cl$ required to decompose $10 g$ of $50%$ pure $C a C O_{3}$ is

$1.825 m L$

$3.65 m L$

$0.9125 m L$

$2 m L$

$(50%$ pure) $5 g \frac{7.3}{2} g H Cl$
$\frac{mass}{volume} =$ density
$∴$ Volume of $H Cl = \frac{mass of H Cl}{density of H Cl} = \frac{7.3}{2 \times 1.825}$
$= 2 m L$

Q. CaCO3​ is decomposed by HCl (density 1.825g/cc ) CaCO3​+2HCl→CaCl2​+H2​O+CO2​ Volume of HCl required to decompose 10g of 50% pure CaCO3​ is

1.825mL

3.65mL

0.9125mL

2mL