Question

$40$ g of carbon is allowed to burn in $56$ litres of oxygen measured at STP. The percentage of unreacted carbon is:

• A. 50%
• B. 75%
• C. 25%
• D. 40%

Answer 1

Answer: (C)

25%

Combustion of carbon is represented by the reaction,

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

40 gms of carbon corresponds to $\frac{40}{12}=3.33$ moles.

56 liters of oxygen at STP corresponds to $\frac{56}{22.4}=2.5$ moles of oxygen.

One mole of carbon reacts with 1 mole of oxygen. Thus, out of 3.33 moles of carbon, 2.5 moles will react with 2.5 moles of oxygen.

Moles of unreacted carbon $=3.33-2.5=0.83moles$

The percentage of unreacted carbon is $\frac{3.33-2.5}{3.33}\times 100=25$%.