Question:
48 g of methane was completely burnt in the presence of oxygen. The liberated gas was passed into a solution containing 370 g of Ca(OH)\(_2\). What is the weight (in g) of CaCO\(_3\) formed?
(Atomic weights: C = 12, H = 1, Ca = 40, O = 16)
48 g of methane was completely burnt in the presence of oxygen. The liberated gas was passed into a solution containing 370 g of Ca(OH)\(_2\). What is the weight (in g) of CaCO\(_3\) formed?
(Atomic weights: C = 12, H = 1, Ca = 40, O = 16)
(Atomic weights: C = 12, H = 1, Ca = 40, O = 16)
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Burning 1 mole of methane gives 1 mole of CO\(_2\), which reacts 1:1 with Ca(OH)\(_2\) to form CaCO\(_3\).
Updated On: Jan 6, 2026
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The Correct Option is C
Approach Solution - 1
Reaction:
\[
\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}
\]
1 mole of CH\(_4\) produces 1 mole of CO\(_2\)
Molar mass of CH\(_4\) = 12 + 4 = 16 g/mol
\[ \text{Moles of CH}_4 = \frac{48}{16} = 3 \Rightarrow 3 \text{ moles of CO}_2 formed \] Now: \[ \text{CO}_2 + \text{Ca(OH)}_2 \rightarrow \text{CaCO}_3 + \text{H}_2\text{O} \] 3 moles of CO\(_2\) → 3 moles of CaCO\(_3\)
Molar mass of CaCO\(_3\) = 40 + 12 + 48 = 100 g/mol \[ \text{Mass of CaCO}_3 = 3 \times 100 = 300~\text{g} \]
Molar mass of CH\(_4\) = 12 + 4 = 16 g/mol
\[ \text{Moles of CH}_4 = \frac{48}{16} = 3 \Rightarrow 3 \text{ moles of CO}_2 formed \] Now: \[ \text{CO}_2 + \text{Ca(OH)}_2 \rightarrow \text{CaCO}_3 + \text{H}_2\text{O} \] 3 moles of CO\(_2\) → 3 moles of CaCO\(_3\)
Molar mass of CaCO\(_3\) = 40 + 12 + 48 = 100 g/mol \[ \text{Mass of CaCO}_3 = 3 \times 100 = 300~\text{g} \]
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Approach Solution -2
48 g of methane was completely burnt in the presence of oxygen. The liberated gas was passed into a solution containing 370 g of Ca(OH)2. What is the weight (in g) of CaCO3 formed?
Step 1: Write the combustion reaction of methane:
\[ \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} \]
From this equation, 1 mole of CH₄ gives 1 mole of CO₂.
Step 2: Calculate moles of CH₄:
Molar mass of CH₄ = 12 + 4 = 16 g/mol
Given mass = 48 g
\[ \text{Moles of CH}_4 = \frac{48}{16} = 3\ \text{mol} \]
So, moles of CO₂ formed = 3 mol
Step 3: Reaction of CO₂ with Ca(OH)₂:
\[ \text{CO}_2 + \text{Ca(OH)}_2 \rightarrow \text{CaCO}_3 + \text{H}_2\text{O} \]
1 mole of CO₂ gives 1 mole of CaCO₃
So 3 mol of CO₂ will give 3 mol of CaCO₃
Step 4: Calculate mass of CaCO₃ formed:
Molar mass of CaCO₃ = 40 (Ca) + 12 (C) + 16×3 (O) = 100 g/mol
\[ \text{Mass of CaCO}_3 = 3 \times 100 = 300\ \text{g} \]
Final Answer:
\[ \boxed{300\ \text{g}} \]
Step 1: Write the combustion reaction of methane:
\[ \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} \]
From this equation, 1 mole of CH₄ gives 1 mole of CO₂.
Step 2: Calculate moles of CH₄:
Molar mass of CH₄ = 12 + 4 = 16 g/mol
Given mass = 48 g
\[ \text{Moles of CH}_4 = \frac{48}{16} = 3\ \text{mol} \]
So, moles of CO₂ formed = 3 mol
Step 3: Reaction of CO₂ with Ca(OH)₂:
\[ \text{CO}_2 + \text{Ca(OH)}_2 \rightarrow \text{CaCO}_3 + \text{H}_2\text{O} \]
1 mole of CO₂ gives 1 mole of CaCO₃
So 3 mol of CO₂ will give 3 mol of CaCO₃
Step 4: Calculate mass of CaCO₃ formed:
Molar mass of CaCO₃ = 40 (Ca) + 12 (C) + 16×3 (O) = 100 g/mol
\[ \text{Mass of CaCO}_3 = 3 \times 100 = 300\ \text{g} \]
Final Answer:
\[ \boxed{300\ \text{g}} \]
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