Question

1.42 g of sodium sulphate is dissolved in 100 mL water. 4.16 g of barium chloride is dissolved in another 100 mL water. The amount of precipitate of barium sulphate upon mixing of the two solutions is:

$N{a}_{2}S{O}_4+BaC{l}_2\to BaS{O}_4+2NaCl$
(Atomic mass: Na = 23, S = 32, O =16, Ba = 137, Cl = 35.5)

• A. 2.33 g
• B. 4.66 g
• C. 2.52 g
• D. 1.42 g

Answer 1

The correct option is A 2.33 g

Molar mass of sodium sulphate $\left(N{a}_{2}S{O}_4\right)=23\times 2+32+16\times 4=142g/mol$

Number of moles of sodium sulphate = $\frac{1.42}{142}=0.01mol$

Molar mass of barium chloride $\left(BaC{l}_2\right)=137+35.5\times 2=137+71=208g/mol$

Number of moles of barium chloride = $\frac{4.16}{208}=0.02mol$

Hence, sodium sulphate is the limiting reagent.

1 mole of sodium sulphate gives 1 mole of barium sulphate.

Hence, 0.01 mole of sodium sulphate gives 0.01 mole of barium sulphate.

Molar mass of barium sulphate $\left(BaS{O}_4\right)=137+32+16\times 4=233g/mol$

Hence, mass of the barium sulphate precipitate formed = $0.01\times 233=2.33g$