2.0 g of mixure of `Na_2CO_3`, `NaHCO_3`, and NaCl on heating produced 56 " mL of " `CO_2` at STP. 1.6g of the same mixture required 25 " mL of " 0.5 M `H_2SO_4` for complete neutralisation. Calculate the percentage of each component present in the mixture.

Explanation

To solve the problem step by step, we will analyze the reactions and use stoichiometry to find the percentages of each component in the mixture of Na₂CO₃, NaHCO₃, and NaCl.

### Step 1: Determine the moles of CO₂ produced
We know that 56 mL of CO₂ is produced at STP. At STP, 1 mole of gas occupies 22.4 L (or 22400 mL).

\[ \text{Moles of CO₂} = \frac{\text{Volume of CO₂ (mL)}}{22400 \text{ mL/mol}} = \frac{56 \text{ mL}}{22400 \text{ mL/mol}} = 0.0025 \text{ moles} \]
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