Question

Combustion of glucose \((C_6H_{12}O_6)\) produces \(CO_2\) and water. The amount of oxygen (in g) required for the complete combustion of 900 g of glucose is: 

[Molar mass of glucose in g \(mol^{-1}\) = 180] 

(1) 480 

(2) 960 

(3) 800 

(4) 32

Answer 1

Correct option is (2) 960 

\(C_6 H_{12} O_{6}+ 6O_{2} \rightarrow 6CO_{2}+ 6 H_2 O\)

Given Mass of glucose = 900 gram 

Molar mass of glucose = 180 gm/mole

So, moles of glucose \(\Rightarrow \frac {900}{180} = 5\)

1 mole of glucose require 6 moles of oxygen gas 

So, 5 moles of glucose will require 30 moles of oxygen gas 

Moles of oxygen gas = 30

Molar mass of oxygen gas \(\Rightarrow\) 32 g/mol 

So, mass of oxygen gas required \(\Rightarrow\) 30 \(\times\) 32 

\(\Rightarrow\) 960 gram.