Question

Mass of sucrose $\left(C_{12}{H}_{22}{O}_{11}\right)$ produced by mixing 84 g of carbon, 12 g of hydrogen and 56 $L$ $O_2$ at 1 atm & 273 K according to given reaction, is $C\left(s\right)+H_2\left(g\right)+O_2\left(g\right)\to C_{12}{H}_{22}{O}_{11}\left(s\right)$

• A. 138.5
• B. 155.5
• C. 172.5
• D. 199.5

Answer 1

The correct option is B 155.5

The balanced reaction, is $12C\left(s\right)+11H_2\left(g\right)+\frac{11}{2}{O}_2\left(g\right)\to C_{12}{H}_{22}{O}_{11}\left(s\right)$
Mole = $\frac{Givenmass}{Molarmass}$
Moles of $C=\frac{84}{12}=7$ mol
Moles of $H_2=\frac{12}{2}=6$ mol
At STP 22.4 $L$ is occupied by 1 mol of the gas.
Using unitary method, 56 $L$ is occupied by $\frac{1}{22.4}\times 56=2.5mol$.
Balanced chemical reaction :
$12C+11H_2+\frac{11}{2}{O}_2\to C_{12}{H}_{22}{O}_{11}$
$\text{For limiting reagent}=\frac{\text{Mole}}{\text{Stoichiometric coefficient}}$
For $C$ = $\frac{7}{12}=0.583$
For $H_2$ = $\frac{6}{11}=0.545$
For $O_2$ = $\frac{2.5}{\left(\frac{11}{2}\right)}=0.454$
Since the ratio is least for $O_2$, hence it is limiting reagent.
According to stoichiometry of reaction, $\frac{11}{2}$ mole $O_2$ produce 1 mole sucrose.
According to unitary method, $\frac{5}{2}$ mol $O_2$ will produce $\frac{2}{11}\times \frac{5}{2}=\frac{5}{11}$ mol of sucrose
The mass of sucrose produced $=\frac{5}{11}\times$(Molar mass of sucrose)
$\Rightarrow \frac{5}{11}\times 342$
$=155.45g$