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JEE Main 2014 · 12 Apr · Shift 1, Chemistry question 34

The amount of BaSOX4\ce{BaSO4} formed upon mixing 100 mL of 20.8% BaClX2\ce{BaCl2} solution with 50 mL of 9.8% HX2SOX4\ce{H2SO4} solution will be :

(Ba = 137, Cl = 35.5, S = 32, H = 1 and O = 16)

Choose one option
Answer and worked solution

Official answer: B

  • BaClX2\ce{BaCl2}: 20.8 g in 100 mL, i.e. 20.8208=0.1\frac{20.8}{208}=0.1 mol.
  • HX2SOX4\ce{H2SO4}: 4.9 g in 50 mL, i.e. 4.998=0.05\frac{4.9}{98}=0.05 mol (limiting).

BaSOX4\ce{BaSO4}: 0.05×233=11.650.05\times233=11.65 g.

Topics:StoichiometryConcentration terms
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