Question

If $27\text{g}$ of carbon is mixed with $88\text{g}$ of oxygen and is allowed to burn to produce $C{O}_2$, then:

• A. Oxygen is the limiting reagent
• B. Volume of $C{O}_2$ gas produced at NTP is $50.4\text{L}$
• C. $C$ and $O$ combine in mass ratio $3:8$
• D. Volume of unreacted $O_2$ at STP is $11.2\text{L}$

Answer 1

Answer: (B), (C), (D)

The correct options are
B Volume of $C{O}_2$ gas produced at NTP is $50.4\text{L}$
C $C$ and $O$ combine in mass ratio $3:8$
D Volume of unreacted $O_2$ at STP is $11.2\text{L}$

$\begin{array}{cccccc}& C& +& O_2& \to & C{O}_2\\ \text{mass}& 27& & 88& & \\ \text{moles}& \frac{27}{12}=2.25& & \frac{88}{32}=2.75& & \end{array}$

therefore $C$ is limiting reagent
Moles of $C{O}_2$ produced = moles of $C$
$=\frac{27}{12}=2.25$
$\therefore$ Volume of $C{O}_2$ at STP $=2.25\times 22.4$
$=50.4\text{L}$

Ratio of masses of $C$ and $O$ in $C{O}_2=12:32=3:8$

Moles of unreacted $O_2=2.75-2.25=0.5$
$\therefore$ Volume of unreacted $O_2$ at STP
$=0.5\times 22.4=11.2\text{L}$