Question

For the combustion of sucrose:
$C_{12}{H}_{22}{O}_{11}+12O_2\to 12C{O}_2+11H_{2}O$
10.0 g of sucrose and 10.0 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?

• A. Sucrose; 5.02 g
• B. Sucrose; 2.53 g
• C. Oxygen; 1.09 g
• D. Oxygen; 5.02 g

Answer 1

The correct option is C Oxygen; 1.09 g

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