Question

A mixture contains N a H C O 3 together with unreactive components. A 1.56 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

Answer 1

Answer:

$\% NaHCO_3=68.6\%$

Explanation:

Hello!

In this case, since a symbolic representation of the undergoing chemical reaction is:

$NaHCO_3 + HA \rightarrow NaA + CO_2 + H_2O$

In such a way, since there is 1:1 mole ratio between the sodium bicarbonate (84.01 g/mol) and carbon dioxide (44.01 g/mol) we compute the mass of reactant that was used during the reaction via stoichiometry:

$m_{NaHCO_3}=0.561gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molNaHCO_3}{1molCO_2}*\frac{84.01gNaHCO_3}{1molNaHCO_3}\\\\m_{NaHCO_3}=1.07gNaHCO_3$

Thus, the by-mass percent of pure sodium bicarbonate in the 1.56-g sample is:

$\% NaHCO_3=\frac{1.07g}{1.56g}*100\%\\\\ \% NaHCO_3=68.6\%$

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