Question

If of conservation of mass was to hold true, then 20.8 g of ${BaCl}_2$ on reaction with 9.8 g of $H_2{SO}_4$ will produce 7.3 g of $HCl$ and ${BaSO}_4$ equal to :

• A. 11.65 g
• B. 23.3 g
• C. 25.5 g
• D. 30.6 g

Answer 1

Answer: (B)

23.3 g

$BaC{l}_2+H_{2}S{O}_4⟶BaS{O}_4+2HCl$

$1$ mole of $BaC{l}_2$ reacts with $1$ mole of $H_{2}S{O}_4$ to give $1$ mole of $BaS{O}_4$ and $2$ moles of $HCl$.

Here, moles of $BaC{l}_2=\frac{20.8}{208}=$ moles of $H_{2}S{O}_4=\frac{9.8}{98}=0.1$

$\therefore$ Moles of $BaS{O}_4$ formed $=0.1$

$\therefore$ Mass of $BaS{O}_4$ formed $=0.1\times 233=23.3g$

i.e. $20.8+9.8=7.3+23.3=30.6$

Hence the correct option is B.