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 \(900 \, \text{g}\) of glucose is:
\([ \text{Molar mass of glucose in g mol}^{-1} = 180 ]\)
\([ \text{Molar mass of glucose in g mol}^{-1} = 180 ]\)
- 480
- 960
- 800
- 32
The Correct Option is B
Approach Solution - 1
C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O
Given Mass of glucose = 900 gram
Molar mass of glucose = 180 gm/mole
So, moles of glucose:
moles of glucose ⇒ 900/180 = 5
1 mole of glucose requires 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 = 32 g/mole
So, mass of oxygen gas required:
mass of oxygen gas required ⇒ 30 × 32 = 960 gram
Approach Solution -2
The balanced combustion reaction of glucose is:
\[\text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2 \rightarrow 6\text{CO}_2 + 6\text{H}_2\text{O}.\]
From the equation:
1 mol of glucose requires 6 mol of $\text{O}_2$.
Molar mass of glucose = $180 \, \text{g/mol}$.
Molar mass of $\text{O}_2 = 32 \, \text{g/mol}$.
Number of moles of glucose in $900 \, \text{g}$:
\[n = \frac{900}{180} = 5 \, \text{mol}.\]
Oxygen required:
\[\text{Mass of } \text{O}_2 = 5 \cdot 6 \cdot 32 = 960 \, \text{g}.\]
Final Answer:
$960 \, \text{g}$.
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