Question

PLZ help!! i m hlf wy n stck!
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?

Answer 1

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 Na2CO3
Molar mass of NaHCO3 = 1 × 23 + 1 × 1 × 12 + 3 × 16
= 84 g mol−1
Number of moles of NaHCO3
According to the question,

⇒ 84x = 106 − 106x
⇒ 190x = 106
⇒ x = 0.5579
Therefore, number of moles of Na2CO3
= 0.0053 mol
And, number of moles of NaHCO3
= 0.0053 mol
HCl reacts with Na2CO3 and NaHCO3 according to the following equation.

Answer 2

now wht?????