Fer the Free Respense you may ase a colculeter an 1 Pcrlodle TaNe. produciog calcium chlorids, water, and carbon dioxide. (s) Wrise a balancel equation for the rastion. $$ \begin{array}{l} \left.\left.\mathrm{Ca}_{2} \mathrm{O}_{3}(\Omega)+\mathrm{HCl}_{(s)}\right) \longrightarrow \mathrm{CaCl}_{2}(4)\right)+\mathrm{H}_{2} \mathrm{OCO}+\mathrm{CO}_{2}(\mathrm{~s}) \ \text { Aas: } \mathrm{CaCO}_{3}(\theta)+2 \mathrm{HCI}(\mathrm{O}) \longrightarrow \mathrm{CaCl}_{2}(4 \rho)+\mathrm{H}_{2} \mathrm{O}(\rho)+\mathrm{CO} \text { (s } \end{array} $$ (b) When the rcaction aas cornglets, 1.55 g of carbcn diovide gas was collocted. Hrer many moles of calrium cubceate were consemed ia the resction? (c) If all Cocalcium carbocate initially presect in the kmple was curnamed ia the reaction, what pervent by cuss of the sass ple was \&us ss eakines carbocate? (d) If be inert enterial was only silicon dioxile, what sas the mole fncticn of silicon diovide is the mikture? $$ \text { mole fructicn }=N_{6}=\frac{n_{1}}{n_{2 n}} $$ or have eo etteat Juatify your sesponse.

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### Part (a): Write a balanced equation for the reaction.

The reaction involves calcium carbonate ($$ \text{CaCO}_3 $$) reacting with hydrochloric acid ($$ \text{HCl} $$) to produce calcium chloride ($$ \text{CaCl}_2 $$), water ($$ \text{H}_2\text{O} $$), and carbon dioxide ($$ \text{CO}_2 $$).

The unbalanced equation is:
$$
\text{CaCO}_3 + \text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2
$$

To balance the equation, we need to ensure that the number of atoms of each element is the same on both sides. The balanced equation is:
$$
\text{CaCO}_3 + 2 \text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2
$$

### Part (b): Calculate the moles of calcium carbonate consumed in the reaction.

Given that 1.55 g of carbon dioxide gas ($$ \text{CO}_2 $$) was collected, we can calculate the number of moles of $$ \text{CO}_2 $$ produced using its molar mass.

The molar mass of $$ \text{CO}_2 $$ is approximately $$ 44.01 \, \text{g/mol} $$.

Calculating the moles of $$ \text{CO}_2 $$:
$$
n_{\text{CO}_2} = \frac{m_{\text{CO}_2}}{M_{\text{CO}_2}} = \frac{1.55 \, \text{g}}{44.01 \, \text{g/mol}}
$$

Now, let's calculate this value.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{1.55}{44.01}$$
- step1: Convert the expressions:
  $$\frac{\frac{31}{20}}{\frac{4401}{100}}$$
- step2: Multiply by the reciprocal:
  $$\frac{31}{20}\times \frac{100}{4401}$$
- step3: Reduce the numbers:
  $$31\times \frac{5}{4401}$$
- step4: Multiply:
  $$\frac{31\times 5}{4401}$$
- step5: Multiply:
  $$\frac{155}{4401}$$
The number of moles of carbon dioxide ($$ \text{CO}_2 $$) produced is approximately $$ 0.0352 \, \text{mol} $$.

From the balanced equation:
$$
\text{CaCO}_3 + 2 \text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2
$$
we see that 1 mole of $$ \text{CaCO}_3 $$ produces 1 mole of $$ \text{CO}_2 $$. Therefore, the moles of calcium carbonate consumed is also $$ 0.0352 \, \text{mol} $$.

### Part (c): Calculate the percent yield of calcium carbonate.

To find the percent yield, we need to know the initial amount of calcium carbonate present. However, since the problem states that all calcium carbonate was consumed, we can assume that the theoretical yield is equal to the amount consumed.

The percent yield can be calculated using the formula:
$$
\text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100
$$

Since the actual yield is the amount of $$ \text{CO}_2 $$ produced, and the theoretical yield is based on the initial amount of $$ \text{CaCO}_3 $$ consumed, we can express it as:
$$
\text{Percent Yield} = \left( \frac{0.0352 \, \text{mol}}{0.0352 \, \text{mol}} \right) \times 100 = 100\%
$$

### Part (d): Calculate the mole fraction of silicon dioxide in the mixture.

To calculate the mole fraction of silicon dioxide ($$ \text{SiO}_2 $$), we need to know the total number of moles in the mixture. Assuming that the only inert material is silicon dioxide, we can denote:
- $$ n_1 $$ = moles of $$ \text{SiO}_2 $$
- $$ n_2 $$ = moles of $$ \text{CaCO}_3 $$ consumed = $$ 0.0352 \, \text{mol} $$

The total number of moles in the mixture is:
$$
n_{\text{total}} = n_1 + n_2
$$

The mole fraction of silicon dioxide is given by:
$$
\text{Mole Fraction} = \frac{n_1}{n_{\text{total}}}
$$

Since we don't have the value of $$ n_1 $$ (the moles of $$ \text{SiO}_2 $$), we cannot calculate the exact mole fraction without additional information. If you provide the amount of silicon dioxide present, I can calculate the mole fraction for you.
(a) The balanced equation is: $$ \text{CaCO}_3 + 2 \text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2 $$ (b) 0.0352 moles of calcium carbonate were consumed. (c) The percent yield of calcium carbonate is 100%. (d) The mole fraction of silicon dioxide in the mixture cannot be determined without knowing the amount of $$ \text{SiO}_2 $$ present.

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