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

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%.

b

explanation:

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