Question:
x g of methane was burnt completely in the presence of oxygen. The liberated gases were passed into a solution containing 370 g of Ca(OH)$_2$. The weight of white precipitate obtained was 500 g. What is the value of x (in g)? (Given : C = 12; H = 1; Ca = 40; O = 16 u)
x g of methane was burnt completely in the presence of oxygen. The liberated gases were passed into a solution containing 370 g of Ca(OH)$_2$. The weight of white precipitate obtained was 500 g. What is the value of x (in g)? (Given : C = 12; H = 1; Ca = 40; O = 16 u)
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In stoichiometry problems, always start with balanced chemical equations. The mole ratio between reactants and products is crucial for solving these problems. Remember to use molar masses to convert between mass and moles.
Updated On: Jan 6, 2026
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The Correct Option is B
Solution and Explanation
Step 1: Balanced chemical equations.
Combustion of methane: $\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(g)$ Reaction of carbon dioxide with calcium hydroxide: $\text{CO}_2(g) + \text{Ca(OH)}_2(aq) \rightarrow \text{CaCO}_3(s) + \text{H}_2\text{O}(l)$
Step 2: Molar mass of calcium carbonate (CaCO$_3$).
Molar mass = 40 (Ca) + 12 (C) + 3 × 16 (O) = 40 + 12 + 48 = 100 g/mol
Step 3: Moles of calcium carbonate precipitate.
Moles of CaCO$_3 = \frac{500 \text{ g}}{100 \text{ g/mol}} = 5$ moles
Step 4: Moles of carbon dioxide produced.
From the stoichiometry of the second reaction, moles of CO$_2$ = moles of CaCO$_3 = 5$ moles.
Step 5: Moles of methane burnt.
From the stoichiometry of the first reaction, moles of CH$_4$ = moles of CO$_2 = 5$ moles.
Step 6: Mass of methane burnt (x).
Molar mass of methane (CH$_4$) = 12 (C) + 4 × 1 (H) = 16 g/mol Mass of CH$_4$ (x) = moles × molar mass = 5 moles × 16 g/mol = 80 g Final Answer: The final answer is $\boxed{80}$
Combustion of methane: $\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(g)$ Reaction of carbon dioxide with calcium hydroxide: $\text{CO}_2(g) + \text{Ca(OH)}_2(aq) \rightarrow \text{CaCO}_3(s) + \text{H}_2\text{O}(l)$
Step 2: Molar mass of calcium carbonate (CaCO$_3$).
Molar mass = 40 (Ca) + 12 (C) + 3 × 16 (O) = 40 + 12 + 48 = 100 g/mol
Step 3: Moles of calcium carbonate precipitate.
Moles of CaCO$_3 = \frac{500 \text{ g}}{100 \text{ g/mol}} = 5$ moles
Step 4: Moles of carbon dioxide produced.
From the stoichiometry of the second reaction, moles of CO$_2$ = moles of CaCO$_3 = 5$ moles.
Step 5: Moles of methane burnt.
From the stoichiometry of the first reaction, moles of CH$_4$ = moles of CO$_2 = 5$ moles.
Step 6: Mass of methane burnt (x).
Molar mass of methane (CH$_4$) = 12 (C) + 4 × 1 (H) = 16 g/mol Mass of CH$_4$ (x) = moles × molar mass = 5 moles × 16 g/mol = 80 g Final Answer: The final answer is $\boxed{80}$
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