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Question

The amount of BaSO4 formed upon mixing 100mL of 20.8% BaCl2 solution with 50mL of 9.8% H2SO4 solution will be:
[Ba=137, Cl=35.5, S=32, H=1 and O=16].

A
23.3g
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B
11.8g
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C
30.6g
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D
33.2g
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Solution

The correct option is A 11.8g

H2SO4+BaCl2BaSO4+2HCl

W/W 20.8 solution 20.8 g BaSO489.2 g water

In 89.2 g water, 20.8 g BaSO4

In 100 g(100 mL) water, x g BaSO4

x=100×20.889.2=23.32 g BaSO4

Number of moles of BaSO4=23.32Molar mass of BaSO4=23.32233=0.1 moles

W/W 9.8 solution 9.8 g H2SO490.2 g H2O

In 90.2 g water, 9.8 g H2SO4

In 50 g(50 mL) water, y g H2SO4

y=50×9.890.2=5.43 g H2SO4

Number of moles of H2SO4=5.43molar mass of H2SO4=5.4398=0.055 moles

Since H2SO4 is in lesser quantity, it will act as a limiting reagent.

0.055 moles of H2SO4 will react with 0.055 moles of BaCl2 to give 0.055 moles of BaSO4 according to balanced reaction equation.

0.055 moles of BaSO4=z grams BaSO4

1 mole BaSO4=233 g

z=233×0.0551=11.8 g


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