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Question

2A+3B4C+5D
Starting with 7 moles of A, 6 moles of B. Find the moles of C and D formed respectively.

A

8 mol,10 mol

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B

14 mol,17.5 mol

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C

9 mol,11.25 mol

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D

12 mol,15 mol

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Solution

The correct option is A

8 mol,10 mol


2A+3B4C+5D
Dividing the given moles of reactant by the respective stoichiometric coefficient of that reactant we get,
A =72=3.5 and B=63=2
Therefore, B is the limiting reagent.
Hence using the law of stoichiometry the moles of C formed =(4×6)3=8 mol
Similarly, moles of D formed =(5×6)3=10 mol


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