Question

50 mL solution of ${\text{BaCl}}_{\text{2}}$(20.8% w/v) and 100 mL solution of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ (9.8% w/v) are mixed then maximum mass of ${\text{BaSO}}_{\text{4}}$ formed is:

$\left[{\text{BaCl}}_{\text{2}}{\text{+ H}}_{\text{2}}{\text{SO}}_{\text{4}}\to {\text{BaSO}}_{\text{4}}\text{+ 2HCl}\right]$

• A. 23.3g
• B. 46.6g
• C. 29.8g
• D. 11.65g

Answer 1

The correct option is D 11.65g

Mass of $BaC{l}_2=50\times \frac{20.8}{100}=10.4$ $gm$

Mass of $H_{2}S{O}_4=100\times \frac{9.8}{100}=9.8$ $gm$

Moles of $BaC{l}_2=\frac{10.4}{208}=0.05$ $moles$

Moles of $H_{2}S{O}_4=\frac{9.8}{9.8}=0.1$ $moles$
$BaC{l}_2+H_{2}S{O}_4⟶BaS{O}_4+2HCl$

$0.05$ $0.1$

$0.05$ $0.05$ $0.05$

$BaC{l}_2$ is limiting, so moles of $BaS{O}_4$ formed $=0.05$

Mass of $BaS{O}_4=0.05\times 233=11.65$ $g$