4g mixture of sodium carbonate and sodium chloride reacts completely with 50 mili equivalent of HCL. Calculate the percentage of sodium carbonate
Answers
Answer:
Let the mixture contains x g of sodium carbonate and 1−x g of sodium bicarbonate.
The molar masses of sodium carbonate and sodium bicarbonate are 106 g/mol and 84 g/mol respectively
The number of moles of sodium carbonate and sodium bicarbonates are
106
x
and
84
1−x
respectively.
Since, it is an equilmolar mixture,
106
x
=
84
1−x
84x=106−106x
190x=106
x=0.5579
Number of moles of sodium carbonate =
106
0.5579
=0.005263
Number of moles of sodium hydrogen carbonate =
84
1−0.5579
=0.005263
One mole of sodium carbonate will react with 2 moles of HCl and 1 mole of sodium bicarbonate will react with 1 mole of HCl.
Total number of moles of HCl that will completely neutralize the mixture =2×0.005263+0.005263=0.01578 moles
Volume of 0.1 M HCl required =
0.1
0.01578
=0.158L=158 mL.
Explanation:
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