Question

Chemistry question

Content - Quality​

Answer 1

Answer:

The yield of carbonic acid H₂CO₃ is approximately 90.7%.

Explanation:

Relative atomic mass data from a modern periodic table:

• Na: 22.990,
• H: 1.008,
• C: 12.011,
• O: 15.999.

Two formula units of sodium bicarbonate $\mathrm{NaHCO_3}$ decomposes to produce one $\mathrm{H_2 O}$ molecule and one $\mathrm{CO}_2$ molecule.

$\rm 2\;NaHCO_3 \to Na_2CO_3 + H_2 O\; (g) + CO_2\; (g)$.

One $\mathrm{H_2O}$ molecule combines with one $\mathrm{CO}_2$ molecule to produce one $\mathrm{H_2CO_3}$ molecule.

$\rm H_2O + CO_2 \to H_2CO_3$.

Combine the two equations:

$\rm 2\;NaHCO_3 \to Na_2CO_3 + H_2CO_3$.

Overall, two formula units of $\mathrm{NaHCO_3}$ thermally decompose to produce one $\mathrm{H_2CO_3}$ molecule.

What's the theoretical yield of this experiment?

To answer this question, start by finding the number of moles of formula units in 2.36 grams of $\textbf{NaHCO_3}$.

$M(\mathrm{NaHCO_3}) = 22.990 + 1.008 + 12.011 + 3\times 15.999 = \rm 84.006\;g\cdot mol^{-1}$.

$\displaystyle n(\mathrm{NaHCO_3}) = \frac{m}{M} = \rm \frac{2.36\;g}{84.006\;g\cdot mol^{-1}} = 0.0280932\; mol$.

$\rm 0.0280932\; mol$ of $\mathrm{NaHCO_3}$ will make half as many moles of $\mathrm{H_2CO_3}$ molecules. In other words,

$\displaystyle n(\mathrm{H_2CO_3})= \frac{1}{2} \; n(\mathrm{NaHCO_3}) = \rm \frac{1}{2}\times 0.0280932\; mol = 0.140466\; mol$.

What will be the mass of that $0.140466\; mol$ of $\mathrm{H_2CO_3}$ molecules?

$M(\mathrm{H_2CO_3}) = \rm 2\times 1.008 + 12.011 + 3\times 15.999 = 62.024\; g\cdot mol^{-1}$.

$\begin{aligned}m(\mathrm{H_2CO_3}\text{, theoretical yield}) &= n \cdot M\\ &= \rm 0.140466\; mol \times 62.024\; g\cdot mol^{-1}\\&=\rm 0.871227\;g\end{aligned}$.

What's the actual yield of this experiment?

The mass of the solid decreased by $2.36 - 1.57 = \rm 0.79\; g$ in this experiment. However, mass is conserved in chemical reactions. The missing $\rm 0.79\; g$ must be the mass of the gaseous $\rm H_2O$ and $\rm CO_2$ that have escaped from the container.

Similarly, assume that all $\rm H_2O$ and $\rm CO_2$ from the first reaction are converted to carbonic acid. The mass of the carbonic acid produced shall be the same as the sum of the mass of $\rm H_2O$ and $\rm CO_2$. In other words, the actual yield of carbonic acid is $\rm 0.79\; g$.

$\displaystyle \begin{aligned}\text{Percentage Yield} &= \frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times 100\%\\&=\rm \frac{0.79\; g}{0.871227\; g}\times 100\%\\&=\rm 90.7\%\end{aligned}$

Answer 2

The yield of carbonic acid H₂CO₃ is approximately 90.7%.

Explanation:

Relative atomic mass data from a modern periodic table:

Na: 22.990,

H: 1.008,

C: 12.011,

O: 15.999.

Two formula units of sodium bicarbonate decomposes to produce one molecule and one molecule.

.

One molecule combines with one molecule to produce one molecule.

.

Combine the two equations:

.

Overall, two formula units of thermally decompose to produce one molecule.

What's the theoretical yield of this experiment?

To answer this question, start by finding the number of moles of formula units in 2.36 grams of .

.

.

of will make half as many moles of molecules. In other words,

.

What will be the mass of that of molecules?

.

.

What's the actual yield of this experiment?

The mass of the solid decreased by in this experiment. However, mass is conserved in chemical reactions. The missing must be the mass of the gaseous and that have escaped from the container.

Similarly, assume that all and from the first reaction are converted to carbonic acid. The mass of the carbonic acid produced shall be the same as the sum of the mass of and . In other words, the actual yield of carbonic acid is .