Question

Mass of sucrose $C_{12}{H}_{22}{O}_{11}$ produced by mixing 84 g of carbon, 12 g of hydrogen and 56 L $O_2$ at 1 atm and 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 g
• B. 155.5 g
• C. 172.5 g
• D. 199.5 g

Answer 1

The correct option is B 155.5 g

$\text{Number of moles}=\frac{\text{given mass}}{\text{Molar mass}}$
$\text{Moles of}C=\frac{84}{12}=7\text{mol}$
$\text{Moles of}{H}_2=\frac{12}{2}=6\text{mol}$
$\text{Moles of}{O}_2=\frac{56}{22.4}=2.5\text{mol}$
The balanced equation for the reaction,
$12C_{\left(s\right)}+11H_{2\left(g\right)}+\frac{11}{2}{O}_{2\left(g\right)}\to C_{12}{H}_{22}{O}_{11\left(s\right)}$
Finding the limiting reagent:
$\frac{\text{moles}}{\text{stoichiometric coefficient}}=\frac{\text{moles of C}}{12}=\frac{7}{12}$
$\frac{\text{moles of}{H}_2}{11}=\frac{6}{11}$
$\frac{\text{moles of}{O}_2}{5.5}=\frac{2.5}{5.5}$

Hence the limiting reagent is $O_2$
$\frac{11}{2}$ mole $O_2$ produce 1 mole sucrose
$\frac{5}{2}$ mole $O_2$ wil form $\frac{5}{11}$ mole sucrose
mass of sucrose = $\frac{5}{11}\times \text{molar mass}$
$=\frac{5}{11}\times 342$
$=155.45$ g