A mixture contains NaHCO3 together with unreactive components. A 1.54 g sample of the mixture reacts with HA to produce 0.561 g of CO2. What is the percent by mass of NaHCO3 in the original mixture

69.55 (w/w) %

Explanation:

When NaHCO3 reacts with an acid HA, the reaction that occurs is:

NaHCO3 + HA → H2O + NaA + CO2

Where 1 mole of NaHCO3 produce 1 mole of CO2

Thus, we need to convert the mass of CO2 to moles using its molar mass (44g/mol). Then, based on the chemical equation, moles of CO2 produced are equal to moles of NaHCO3 in the mixture. With its molar mass -84g/mol- we can find the mass of NaHCO3 and mass percent:

Moles CO2:

0.561g * (1mol / 44g) = 0.01275 moles CO2 = Moles NaHCO3.

Mass NaHCO3:

0.01275 moles * (84g/mol) = 1.071g NaHCO3

Mass percent:

1.071g NaHCO3 / 1.54g sample * 100

that's just how they are

explanation:

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i know insulin is primary but i dont know the others im sorry

explanation: