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Mixture of MgCO(3) &NaHCO(3) on strong h...

Mixture of `MgCO_(3)` &`NaHCO_(3)` on strong heating gives `CO_(2)`
&`H_(2)O` in 3 : 1 mole ratio. The weight % of `NaHCO_(3)` present in the mixture is :

A

0.3

B

0.8

C

0.4

D

0.5

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The correct Answer is:
To solve the problem, we need to analyze the decomposition reactions of sodium bicarbonate (NaHCO₃) and magnesium carbonate (MgCO₃) when heated, and then calculate the weight percentage of NaHCO₃ in the mixture based on the products formed. ### Step-by-Step Solution: 1. **Identify the Decomposition Reactions:** - Sodium bicarbonate decomposes as follows: \[ 2 \text{NaHCO}_3 \rightarrow \text{Na}_2\text{O} + 2 \text{CO}_2 + \text{H}_2\text{O} \] - Magnesium carbonate decomposes as follows: \[ \text{MgCO}_3 \rightarrow \text{MgO} + \text{CO}_2 \] 2. **Determine the Moles of Products:** - From the sodium bicarbonate reaction, 2 moles of NaHCO₃ produce 2 moles of CO₂ and 1 mole of H₂O. - From the magnesium carbonate reaction, 1 mole of MgCO₃ produces 1 mole of CO₂. - Therefore, the total moles of CO₂ produced from both reactions can be expressed in terms of x (moles of NaHCO₃) and y (moles of MgCO₃): - Total CO₂ = 2x (from NaHCO₃) + y (from MgCO₃) - Total H₂O = x (from NaHCO₃) 3. **Set Up the Mole Ratio:** - According to the problem, the ratio of CO₂ to H₂O is given as 3:1. Therefore, we can write: \[ \frac{2x + y}{x} = 3 \] 4. **Solve for y in terms of x:** - Cross-multiplying gives: \[ 2x + y = 3x \] - Rearranging gives: \[ y = 3x - 2x = x \] 5. **Calculate the Weight Percent of NaHCO₃:** - The molar mass of NaHCO₃ is 84 g/mol, and the molar mass of MgCO₃ is 100 g/mol. - The total mass of the mixture can be expressed as: \[ \text{Total mass} = 84x + 100y = 84x + 100x = 184x \] - The weight percentage of NaHCO₃ in the mixture is given by: \[ \text{Weight \% of NaHCO}_3 = \left(\frac{84x}{184x}\right) \times 100 \] - Simplifying this gives: \[ \text{Weight \% of NaHCO}_3 = \left(\frac{84}{184}\right) \times 100 \approx 45.65\% \] 6. **Final Answer:** - The weight percentage of NaHCO₃ present in the mixture is approximately 45.65%, which can be rounded to 46%.
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