Question

When NaHCO3 completely decomposes, it can follow this balanced chemical equation:2NaHCO3 → Na2CO3 + H2CO3Determine the theoretical yields of each product using stoichiometry if the mass of the NaHCO3 sample is 3.55 grams. (Show work for both)In an actual decomposition of NaHCO3, the mass of one of the products was measured to be 2.16 grams. Identify which product this could be and justify your reasoning.Calculate the percent yield of the product identified in part B. (Show your work)

Answer 1

A) Theoretical yield of Na2CO3 = 2.24grams

Theoretical yield of H2CO3 = 1.31grams

B) Na2CO3

C) 96.43%

Explanations:

A) Given the chemical reaction involving the actual decomposition of NaHCO3 as shown:

`2NaHCO_3\rightarrow Na_2CO_3+H_2CO_3`

Given the following

mass of NaHCO3 = 3.55 grams

Determine the mole of NaHCO3

`\begin{gathered} mole=\frac{mass}{molar\text{ mass}} \\ mole=(3.55)/(84) \\ molar\text{ }mass\text{ }of\text{ }NaHCO₃=0.04226moles \end{gathered}`

Since 2 moles of the reactant produces 1 mole each of the product, the moles of the product will be expressed as:

`\begin{gathered} mole\text{ of }Na_2CO_3=mole\text{ of }H_2CO_3=(0.04226)/(2) \\ mole\text{ of }Na_2CO_3=mole\text{ of }H_2CO_3=0.02113moles \end{gathered}`

Determine the mass of H2CO3 produced (theoretical yield)

`\begin{gathered} mass\text{ of H}_2CO_3=mole* molar\text{ }mass \\ mass\text{ of H}_2CO_3=0.02113*62.03 \\ mass\text{ of H}_2CO_3=1.31grams \end{gathered}`

Determine the mass of Na2CO3 (theoretical yield)

`\begin{gathered} mass\text{ of Na}_2CO_3=0.02113*105.9888 \\ mass\text{ of Na}_2CO_3=2.24grams \end{gathered}`

B) According to the theoretical yields, you can see the product that could yield the mass of 2.16 grams is Na2CO3 since it has the closest mass (2.24g) to the actual mass.

C) Determine the percent yield

`\begin{gathered} \%yield=(actual)/(theoretical)*100 \\ \%yield=(2.16)/(2.24)*100 \\ \%yield=96.43\% \end{gathered}`