How many ml of 0.1M HCL are required to react completely with 1g mixture of Na2CO3 and NaHCO3 containing equimolar amount of two?
just for fun
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Answers
Answer:
Na2Co3 and NaHCO3 are both equimolar. It is the meaning of substances to have the same molarity
Na2CO3 in the mixture with a total amount will be x gram
NaHCO3 in the mixture with a total amount will be (1-x) gram
Now the molar mass of Na2CO3 = 106g/mol
therefore,
So the nuber of moles is Na2CO3 = x/106mol
molar mass of NaHCO3 = 84 g/mol
Therefore, number of moles of NaHCO3 = 1-x/84 mol
As per the given questions
X/106 = 1-X/54
84x = 106 – 106x
190x = 106
x = 0.5579
After that
number of moles of Na2CO3 = 0.5579/106 mol
= 0.0053mol
And number of moles of NaHCO3 = 1 – 0.5579/84
= 0.0053 mol
Hcl reacts with Na2CO3 and Na2CO3 according to the following equation
2HCl + Na2CO3 ------------> 2NaCl + H2O + CO2
2mol 1mol
and
HCl + NaHCO3 ----------->NaCl + H2O + CO2
1mol 1 mol
From the above reactions,
1 mol of Na2Co3 will react with 2 mol of Hcl
Therefore, 0.00526 mol of Na2Co3 will react with 2x 0.00526 mol of Hcl
similarly 0.00526 mol of NaHCO3 will react with 0.00526 mol of Hcl
Total mol of Hcl required to react with a mixture of NaHCO3 & Na2Co3= 2x 0.00526 + 0.00526= 0.01578 mol.
Explanation: