Question

An impure sample of solid Na₂CO₃ is allowed to react with 0.1755 M HCl.


`Na_2CO_3 + 2HCl \longrightarrow 2NaCl + CO_2 + H_2O`
A 0.2337-g sample of sodium carbonate requires 15.55 mL of HCl solution. What is the purity of the sodium carbonate?

Answer 1

Answer: Thus percentage purity of sodium carbonate is 60.83%.

Step-by-step explanation:

`Na_2CO_3 + 2HCl \longrightarrow 2NaCl + CO_2 + H_2O`

Volume of the HCl solution ,V = 15.55 mL= 0.01555 L

Concentration of HCl solution ,C= 0.1755 M

Moles of HCl in 0.1755 M solution = n

`Molarity=(n)/(V(L))`

`n=C\time V=0.1755 M* 0.01555 L = 2.729* 10^(-3) mol`

According o reaction 2 moles of HCl reacts with 1 mole of sodium carbonate.

Then n moles of HCl will react with:

`(1)/(2)* n=(1)/(2)* 2.729* 10^(-3) mol=1.3645* 10^(-3) mol` of sodium carbonate.

Mass of
`1.364* 10^(-3) mol` of sodium carbonate:

`1.364* 10^(-3) mol* 106 g/mol=0.144584 g`

Mass of the sample given= 0.2377 grams

Mass used up in reaction = 0.144585 g

percentage purity=
`(0.144585)/(0.2337)* 100=60.83\%`

Thus percentage purity of sodium carbonate is 60.83%.