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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:

Na2SO4+BaCl2BaSO4+2NaCl
(Atomic mass: Na = 23, S = 32, O =16, Ba = 137, Cl = 35.5)

A
2.33 g
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B
4.66 g
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C
2.52 g
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D
1.42 g
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Solution

The correct option is A 2.33 g
Molar mass of sodium sulphate (Na2SO4)=23×2+32+16×4=142 g/mol

Number of moles of sodium sulphate = 1.42142=0.01 mol

Molar mass of barium chloride (BaCl2)=137+35.5×2=137+71=208 g/mol

Number of moles of barium chloride = 4.16208=0.02 mol

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 (BaSO4)=137+32+16×4=233 g/mol

Hence, mass of the barium sulphate precipitate formed = 0.01×233=2.33 g

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