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Answer» 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»

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 $\rm 0.79\; g$ .

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

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