Answer:
The given reaction is :
\[\underset{2mol}{\mathop{2A}}\,+\underset{4mol}{\mathop{4B}}\,\to
\underset{3mol}{\mathop{3C}}\,+\underset{4mol}{\mathop{4D}}\,\]
Given moles of A and B are 5 and 6 moles
respectively.
Case
I: Let reactant (A) is completely consumed.
2 mol A \[\equiv \] 3
mol C
\[\therefore \] \[5molA\equiv
\frac{3}{2}\times 5molC\equiv 7.5\,molC\]
Case
II: Let reactant (B) is completely consumed.
4
mol 5 \[\equiv \] 3 mol C
\[\therefore \] \[6\text{ }mol\text{ }B\equiv
\frac{3}{4}\times 6mol\text{ }C\text{ }\equiv \text{ }4.5\text{ }mol\text{ }C\]
Since (B) on complete consumption gives
least amount of product (C) hence (B) will be limiting reagent and the amount
of (C) formed will be 4.S mol.
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