Question

For the combustion of sucrose:
$C_{12}{H}_{22}{O}_{11}+12O_2\to 12C{O}_2+11H_{2}O$
$10.0\text{g}$ of sucrose and $10.0\text{g}$ of oxygen are made to react. Which of the following reactants will be the limiting reagent and what amount of the excess reactant will be left?

Answer 1

$1\text{mole}$ of sucrose reacts with $12\text{moles}$ of oxygen
$342\text{g}$ of sucrose reacts with $(12\times 32)\text{g}$ of oxygen.
So, $10\text{g}$ of sucrose will react with $\frac{12\times 32}{342}\times 10=11.228\text{g of oxygen}$.
Hence, oxygen is the limiting reagent.
$12\times 32\text{g}$ of oxygen reacts with $342\text{g}$ of sucrose.
$10\text{g}$ of oxygen will react with $\frac{342}{12\times 32}\times 10=8.906\text{g}$ sucrose.
$\text{Amount of sucrose left}=10-8.906\text{g}=1.09\text{g}$