In the given scheme of reactions, 10 moles of A is taken, while reagent B, E and H are taken in excess. Find the moles of G formed. ReactionYield of reaction(i) 2A+B→3C+D20%(ii) 2C+E→4F40%(iii) 7H+3F→8G50%
A
1.4 mol
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B
3.2 mol
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C
2.8 mol
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D
8.4 mol
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Solution
The correct option is B 3.2 mol ReactionYield(i) 2A+B→3C+D20%(ii) 2C+E→4F40%(iii) 7H+3F→8G50% Initially, the number of mol of A present = 10 In reaction (i), 2 mol of A reacts to produce 3 mol of C 10 mol of A will produce = 32×10=15 mol of C Yield of reaction = 20 % = Actual yieldTheoretical yield×100% Actual yield obtained = 20100×15 = 3 mol of C In reaction (ii), 2 mol of C produces 4 mol of F 3 mol of C will produce = 42×3 = 6 moles Actual yield obtained = 40100×6 = 2.4 mol of F In reaction (iii), 3 mol of F reacts to produce 8 mol of G 2.4 mol of F will produce = 83×2.4 = 6.4 mol of G Actual yield obtained = 50100×6.4 = 3.2 mol of G