Question

If 20 g of carbon is mixed with 88 g of oxygen, and is allowed to burn to produce $C{O}_2$, then which is incorrect?

• A. Oxygen is the limiting reagent
• B. Volume of $C{O}_2$ gas produced at N.T.P. is 50.4 L
• C. C and O combine in mass ratio $3:8$
• D. Both (A) & (C)

Answer 1

Answer: (A)

Oxygen is the limiting reagent

$C+O_2\to C{O}_2$

According to balanced equation: $12g+32g\to 44g$

According to question: $20g+88g\to ?$

Since, $12g$ of $C$ requires $32g$ of $O_2$

So, $20g$ of $C$ requires $=\frac{20\times 32}{12}grams$ of $O_2$

$=53.33g$

Since $32g$ of $O_2$ requires $12g$ of $C$

So, $88g$ of $O_2$ will require $=\frac{88\times 12}{32}grams$ of $C$.

$\therefore ,$ Since $C$ is less than available that is required with $88g$ of $O_2$

So, $C$ is limiting reagent and it will divide course of reaction.