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Question

50 mL solution of BaCl2(20.8% w/v) and 100 mL solution of H2SO4 (9.8% w/v) are mixed then maximum mass of BaSO4 formed is:

[BaCl2 + H2SO4BaSO4 + 2HCl]

A
23.3g
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B
46.6g
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C
29.8g
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D
11.65g
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Solution

The correct option is D 11.65g
Mass of BaCl2=50×20.8100=10.4 gm
Mass of H2SO4=100×9.8100=9.8 gm
Moles of BaCl2=10.4208=0.05 moles
Moles of H2SO4=9.89.8=0.1 moles
BaCl2+H2SO4BaSO4+2HCl
0.05 0.1
0.05 0.05 0.05
BaCl2 is limiting, so moles of BaSO4 formed =0.05
Mass of BaSO4=0.05×233=11.65 g

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