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

Question

In the reaction, \[ 2\text{Al}(s) + 6\text{HCl}(aq) \rightarrow 2\text{Al}^{3+}(aq) + 6\text{Cl}^-(aq) + 3\text{H}_2(g) \]

Answer 1

Chemical Reaction: C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O

Given: Mass of glucose = 900 g

Molar mass of glucose = 180 g/mol

Calculate moles of glucose:

Moles of glucose = Mass / Molar mass = 900 g / 180 g/mol = 5 mol

Stoichiometric ratio: 1 mole of glucose reacts with 6 moles of oxygen gas.

Therefore, 5 moles of glucose will require (5 * 6) = 30 moles of oxygen gas.

Moles of oxygen gas required = 30 mol

Molar mass of oxygen gas = 32 g/mol

Calculate the mass of oxygen gas required:

Mass of oxygen gas required = Moles × Molar mass = 30 mol × 32 g/mol = 960 g

The Correct Option is B