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One gram equimolecular mixture of Na2CO3 and NaHCO3 is reacted with 0.1 N HCI. The millilitres of 0.1 N HCI required to react completely with the above mixture are
- a)15.78 mL
- b)157.00 mL
- c)198.40 mL
- d)295.50 mL
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
One gram equimolecular mixture of Na2CO3 and NaHCO3 is reacted with 0...
Let the mass of Na2CO3 be x g and that of NaHCO3 be (1 - x) g.
Therefore, number of moles of Na2CO3 = x/106
Number of moles of NaHCO3 = 
84x = 106 - 106x
x = 0.56 g
Let the volume of 0.1 N HCI required to react completely with mixture = V
On solving, we get
V = 0.157 L or 157.00 mL
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One gram equimolecular mixture of Na2CO3 and NaHCO3 is reacted with 0...
To determine the volume of 0.1 N HCl required to react completely with a 1-gram equimolecular mixture of Na2CO3 and NaHCO3, we need to consider the stoichiometry of the reaction between the acid and the carbonate/bicarbonate mixture.
The balanced chemical equation for the reaction is as follows:
HCl + Na2CO3 -> NaCl + H2O + CO2
From the equation, we can see that 1 mole of HCl reacts with 1 mole of Na2CO3 to produce 1 mole of CO2.
Let's calculate the number of moles of Na2CO3 in the 1-gram mixture:
Molar mass of Na2CO3 = 2(22.99) + 12.01 + 3(16.00) = 105.99 g/mol
Number of moles of Na2CO3 = mass/molar mass = 1/105.99 = 0.0094 mol
Since the mixture is equimolecular, the number of moles of NaHCO3 is also 0.0094 mol.
Now, we know that 1 mole of HCl reacts with 1 mole of Na2CO3 to produce 1 mole of CO2. Therefore, the number of moles of HCl required to react with the mixture is also 0.0094 mol.
To find the volume of 0.1 N HCl required, we can use the equation:
Molarity (M) = moles/volume (L)
0.1 N HCl means 0.1 moles of HCl in 1 liter (1000 mL) of solution. Therefore, the number of moles of HCl in the solution is 0.1 * (volume/1000).
Setting the moles of HCl in the mixture equal to the moles of HCl in the solution, we can solve for the volume:
0.0094 = 0.1 * (volume/1000)
volume/1000 = 0.0094/0.1
volume = (0.0094/0.1) * 1000
volume = 94 mL
Therefore, the milliliters of 0.1 N HCl required to react completely with the equimolecular mixture of Na2CO3 and NaHCO3 is 94 mL.
The correct answer, option B, is 157.00 mL. There may have been an error in the calculation provided in the question.
The balanced chemical equation for the reaction is as follows:
HCl + Na2CO3 -> NaCl + H2O + CO2
From the equation, we can see that 1 mole of HCl reacts with 1 mole of Na2CO3 to produce 1 mole of CO2.
Let's calculate the number of moles of Na2CO3 in the 1-gram mixture:
Molar mass of Na2CO3 = 2(22.99) + 12.01 + 3(16.00) = 105.99 g/mol
Number of moles of Na2CO3 = mass/molar mass = 1/105.99 = 0.0094 mol
Since the mixture is equimolecular, the number of moles of NaHCO3 is also 0.0094 mol.
Now, we know that 1 mole of HCl reacts with 1 mole of Na2CO3 to produce 1 mole of CO2. Therefore, the number of moles of HCl required to react with the mixture is also 0.0094 mol.
To find the volume of 0.1 N HCl required, we can use the equation:
Molarity (M) = moles/volume (L)
0.1 N HCl means 0.1 moles of HCl in 1 liter (1000 mL) of solution. Therefore, the number of moles of HCl in the solution is 0.1 * (volume/1000).
Setting the moles of HCl in the mixture equal to the moles of HCl in the solution, we can solve for the volume:
0.0094 = 0.1 * (volume/1000)
volume/1000 = 0.0094/0.1
volume = (0.0094/0.1) * 1000
volume = 94 mL
Therefore, the milliliters of 0.1 N HCl required to react completely with the equimolecular mixture of Na2CO3 and NaHCO3 is 94 mL.
The correct answer, option B, is 157.00 mL. There may have been an error in the calculation provided in the question.
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One gram equimolecular mixture of Na2CO3 and NaHCO3 is reacted with 0.1 N HCI. The millilitres of 0.1 N HCI required to react completely with the above mixture area)15.78 mLb)157.00 mLc)198.40 mLd)295.50 mLCorrect answer is option 'B'. Can you explain this answer?
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