Question

A mixture contains NaHCO3

together with unreactive components. A 1.68 g sample of the mixture reacts with HA
to produce 0.561 g of CO2

.

The molar mass of NaHCO3
is 84.01 g/mol
and the molar mass of CO2
is 44.01 g/mol.
What is the percent by mass of NaHCO3
in the original mixture?

Answer 1

Answer: 63.75 %

Step-by-step explanation:

The balanced chemical equation for this reaction is:

NaHCO3 + HA = NaA + H2O + CO2

According to the chemical equation, one mole of NaHCO3 will react with one mole of HA to form one mole of NaA, H2O and CO2. All the components are in same mole ratio of 1:1:1:1

So,

Moles of CO2 = weight/molar mass = 0.561/44 = 0.01275 moles

Moles of CO2 will be equal to moles of NaHCO3 = 0.01275 moles

Moles = weight/mass

Weight = moles × mass

So the weight of NaHCO3 = 0.01275 × 84.01 = 1.071 g

The percentage mass of NaHCO3 in the original mixture =(1.071/1.68)× 100% = 63.75 %

Therefore, the percent by mass of NaHCO3 in the original mixture was 63.75 %