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प्रश्न
How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na2CO3 and NaHCO3 containing equimolar amounts of both?
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उत्तर
Let the amount of Na2CO3 in the mixture be x g.
Then, the amount of NaHCO3 in the mixture is (1 − x) g.
Molar mass of Na2CO3 = 2 × 23 + 1 × 12 + 3 × 16
= 106 g mol−1
∴ Number of moles of Na2CO3 = `x/106` mol
Molar mass of NaHCO3 = 1 × 23 + 1 × 1 × 12 + 3 × 16
= 84 g mol−1
∴ Number of moles of NaHCO3 = `(1 - x)/84` mol
According to the question,
`x/106 = (1 - x)/84`
⇒ 84x = 106 − 106x
⇒ 190x = 106
⇒ x = `106/190`
⇒ x = 0.558
Therefore, number of moles of Na2CO3 = `0.558/106` mol
= 0.00526 mol
And, number of moles of NaHCO3 = `(1 - 0.558)/84`
= 0.00526 mol
HCl reacts with Na2CO3 and NaHCO3 according to the following equation:
\[\ce{\underset{2 mol}{2 HCl} + \underset{1 mol}{Na2CO3} -> 2 NaCl + H2O + CO2}\]
\[\ce{\underset{1 mol}{HCl} + \underset{1 mol}{NaHCO3} -> NaCl + H2O + CO2}\]
1 mol of Na2CO3 reacts with 2 mol of HCl.
Therefore, 0.00526 mol of Na2CO3 reacts with 2 × 0.00526 mol = 0.01052 mol.
Similarly, 1 mol of NaHCO3 reacts with 1 mol of HCl.
Therefore, 0.00526 mol of NaHCO3 reacts with 0.00526 mol of HCl.
Total moles of HCl required = (0.01052 + 0.00526) mol
= 0.01578 mol
In 0.1 M of HCl,
0.1 mol of HCl is present in 1000 mL of the solution.
Therefore, volume of solution = `(1000 xx 0.01578)/0.1` mL
= 157.8 mL of the solution
Hence, 157.8 mL of 0.1 M HCl is required to react completely with a 1 g mixture of Na2CO3 and NaHCO3, containing equimolar amounts of both.
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