A mixture contains N a H C O 3 together with unreactive components. A 1.68 g sample of the mixture reacts with H A to produce 0.561 g of C O 2 . What is the percent by mass of N a H C O 3 in the original mixture

There is 65% of NaHCO3 in the sample.

The equation of the reaction is;

HA + NaHCO3 -----> NaA + CO2 + H2O

Amount of CO2 formed = mass/molar mass

mass of CO2 = 0.561 g/44 g/mol = 0.013 moles

From the balanced reaction equation;

1 mole of NaHCO3 yields 1 mole of CO2

0.013 moles of Na2CO3 yields 0.013 moles of CO2

Hence, mass of NaHCO3 in the sample = 0.013 moles × 84 g/mol = 1.092 g of NaHCO3

Percentage by mass of NaHCO3 = 1.092 g/1.68 g ×100/1

= 65%

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Answer:

63.75%.

Explanation:

The first step here is to write out the reaction showing the chemical reaction between the two chemical species. Thus, we have;

HA(aq) + NaHCO3 --------------> CO2(g) + H20(l) + NaA(aq).

Therefore, the mole ratio is 1 : 1 : 1 : 1 that is go say one mole of HA reacted with one mole of NaHCO3 to give one mole of CO2 and one .ole of NaA.

Hence, the number of moles of CO2 = mass/molar mass = 0.561/44 = 0.01275 moles.

Thus, the number of moles of NaHCO3 = number of moles of CO2 = 0.01275 moles.

Therefore, we have ( 0.01275 moles × 84 g/mol) grams = 1.071 g NaHCO3 in the mixture.

Therefore, the percent by mass of N a H C O 3 in the original mixture = 1.071/1.68 × 100 = 63.75%.