Answer:
[tex]\% NaHCO_3=68.6\%[/tex]
Explanation:
Hello!
In this case, since a symbolic representation of the undergoing chemical reaction is:
[tex]NaHCO_3 + HA \rightarrow NaA + CO_2 + H_2O[/tex]
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:
[tex]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[/tex]
Thus, the by-mass percent of pure sodium bicarbonate in the 1.56-g sample is:
[tex]\% NaHCO_3=\frac{1.07g}{1.56g}*100\%\\\\ \% NaHCO_3=68.6\%[/tex]
Best regards!
The percent by mass of NaHCO₃ in the original mixture is 68.65%
We'll begin by calculating the mass of NaHCO₃ that reacted and the mass of CO₂ produced from the balanced equation.
NaHCO₃ + HA —> NaA + CO₂ + H₂O
Molar mass of NaHCO₃ = 23 + 1 + 12 + (16×3) = 84 g
Mass of NaHCO₃ from the balanced equation = 1 × 84 = 84 g
Molar mass of CO₂ = 12 + (16×2) = 44 g/mol
Mass of CO₂ from the balanced equation = 1 × 44 = 44 g
From the balanced equation above,
84 g of NaHCO₃ reacted to produce 44 g of CO₂
From the balanced equation above,
84 g of NaHCO₃ reacted to produce 44 g of CO₂
Therefore,
X g of NaHCO₃ will react to produce 0.561 g of CO₂ i.e
[tex]X \: g \: of \: NaHCO_3 = \frac{84 \times 0.561}{44} \\ \\ X \: g \: of \: NaHCO_3 = 1.071 \: g[/tex]
Mass of NaHCO₃ = 1.071 g
Mass of mixture = 1.56 g
[tex]Percentage \: = \frac{mass}{total \: mas} \times 100 \\ \\ = \frac{1.071}{1.56} \times 100 \\ \\ [/tex]
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