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

Explanation:

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

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:

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

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it’s previous models   because when i picked that son of a gun’s answer it was wrong look for previous models because they might make it a different letter