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4.0 g of a mixture of Nacl and Na(2) CO(...

`4.0 g` of a mixture of `Nacl` and `Na_(2) CO_(3)` was dissolved in water and volume made up to `250 mL`. `25 mL` of this solution required `50 mL` of `N//10 HCl` for complete neutralisation. Calculate the percentage composition of the original mixture.

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To solve the problem step by step, we will follow the information given and the chemical reactions involved. ### Step 1: Understand the Reaction The reaction between sodium carbonate (Na₂CO₃) and hydrochloric acid (HCl) is as follows: \[ \text{Na}_2\text{CO}_3 + 2 \text{HCl} \rightarrow 2 \text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 \] This shows that 1 mole of Na₂CO₃ reacts with 2 moles of HCl. ### Step 2: Calculate the Normality of HCl We are given that 50 mL of N/10 HCl is used. Normality (N) is defined as equivalents per liter. Therefore, N/10 means: \[ \text{N} = \frac{1}{10} \text{ equivalents/L} \] For 50 mL: \[ \text{Equivalents of HCl} = \text{Volume (L)} \times \text{Normality} = 0.050 \, \text{L} \times \frac{1}{10} = 0.005 \, \text{equivalents} \] ### Step 3: Relate HCl Equivalents to Na₂CO₃ Since 1 mole of Na₂CO₃ reacts with 2 equivalents of HCl, the equivalents of Na₂CO₃ can be calculated as: \[ \text{Equivalents of Na}_2\text{CO}_3 = \frac{0.005}{2} = 0.0025 \, \text{equivalents} \] ### Step 4: Calculate the Mass of Na₂CO₃ To find the mass of Na₂CO₃, we need to use its molar mass. The molar mass of Na₂CO₃ is: \[ \text{Molar mass of Na}_2\text{CO}_3 = 2(23) + 12 + 3(16) = 106 \, \text{g/mol} \] Now, using the equivalents to find the mass: \[ \text{Mass of Na}_2\text{CO}_3 = \text{Equivalents} \times \text{Molar mass} \] \[ = 0.0025 \times 106 = 0.265 \, \text{g} \] ### Step 5: Calculate the Total Mass of the Mixture We know the total mass of the mixture is 4.0 g. The mass of NaCl can be calculated as: \[ \text{Mass of NaCl} = \text{Total mass} - \text{Mass of Na}_2\text{CO}_3 \] \[ = 4.0 \, \text{g} - 0.265 \, \text{g} = 3.735 \, \text{g} \] ### Step 6: Calculate the Percentage Composition Now, we can find the percentage composition of Na₂CO₃ and NaCl in the mixture: 1. Percentage of Na₂CO₃: \[ \text{Percentage of Na}_2\text{CO}_3 = \left( \frac{0.265}{4.0} \right) \times 100 = 6.625\% \] 2. Percentage of NaCl: \[ \text{Percentage of NaCl} = \left( \frac{3.735}{4.0} \right) \times 100 = 93.375\% \] ### Final Answer - Percentage of Na₂CO₃ = 6.625% - Percentage of NaCl = 93.375%
To solve the problem step by step, we will follow the information given and the chemical reactions involved. ### Step 1: Understand the Reaction The reaction between sodium carbonate (Na₂CO₃) and hydrochloric acid (HCl) is as follows: \[ \text{Na}_2\text{CO}_3 + 2 \text{HCl} \rightarrow 2 \text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 \] This shows that 1 mole of Na₂CO₃ reacts with 2 moles of HCl. ### Step 2: Calculate the Normality of HCl ...
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