Question
For a reaction $$A + 2B \to C,$$ the amount of $$C$$ formed by starting the reaction with $$5\,moles$$ of $$A$$ and $$8\,moles$$ of $$B$$ is
A.
5$$\,moles$$
B.
8$$\,moles$$
C.
16$$\,moles$$
D.
4$$\,moles$$
Answer :
4$$\,moles$$
Solution :
$$A + 2B \to C$$
$$1\,mole$$ of $$A$$ reacts with $$2\,moles$$ of $$B$$ to give $$1\,mole$$ of $$C.$$
∴ $$5\,moles$$ of $$A$$ would react with $$10\,moles$$ of $$B$$ to give $$5\,moles$$ of $$C.$$
But, only $$8\,moles$$ of $$B$$ are available
∴ $$B$$ acts as a limiting reagent.
$$2\,moles$$ of $$B$$ gives $$1\,mole$$ of C
∴ $$8\,moles$$ of $$B$$ will give $$\frac{1}{2} \times 8 = 4\,moles$$ of $$C.$$