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solve the following problems. 6. how many grams of oxygen can be produc…

Question

solve the following problems.

  1. how many grams of oxygen can be produced from the decomposition of 100. g of $\text{kclo}_3$?
  2. how much iron can be recovered from 25.0 g of $\text{fe}_2\text{o}_3$?
  3. how much silver can be produced from 125 g of $\text{ag}_2\text{s}$?

Explanation:

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Problem 6

Step1: Balance the decomposition reaction

The balanced chemical equation for the decomposition of $\text{KClO}_3$ is:

$$2\text{KClO}_3 ightarrow 2\text{KCl} + 3\text{O}_2$$
Step2: Calculate molar mass of $\text{KClO}_3$

Molar mass of $\text{KClO}_3 = 39.10 + 35.45 + 3\times16.00 = 122.55\ \text{g/mol}$

Step3: Find moles of $\text{KClO}_3$

$$\text{Moles of KClO}_3 = \frac{100.\ \text{g}}{122.55\ \text{g/mol}} \approx 0.816\ \text{mol}$$

Step4: Relate moles to $\text{O}_2$

From the equation, 2 mol $\text{KClO}_3$ produces 3 mol $\text{O}_2$.
$$\text{Moles of O}_2 = 0.816\ \text{mol} \times \frac{3}{2} = 1.224\ \text{mol}$$

Step5: Calculate mass of $\text{O}_2$

Molar mass of $\text{O}_2 = 2\times16.00 = 32.00\ \text{g/mol}$
$$\text{Mass of O}_2 = 1.224\ \text{mol} \times 32.00\ \text{g/mol} \approx 39.2\ \text{g}$$
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Problem 7

Step1: Balance the reduction reaction

The reaction for recovering Fe from $\text{Fe}_2\text{O}_3$ (simplified):

$$\text{Fe}_2\text{O}_3 ightarrow 2\text{Fe} + \frac{3}{2}\text{O}_2$$
Step2: Calculate molar mass of $\text{Fe}_2\text{O}_3$

Molar mass of $\text{Fe}_2\text{O}_3 = 2\times55.85 + 3\times16.00 = 159.70\ \text{g/mol}$

Step3: Find moles of $\text{Fe}_2\text{O}_3$

$$\text{Moles of Fe}_2\text{O}_3 = \frac{25.0\ \text{g}}{159.70\ \text{g/mol}} \approx 0.1566\ \text{mol}$$

Step4: Relate moles to Fe

1 mol $\text{Fe}_2\text{O}_3$ produces 2 mol Fe.
$$\text{Moles of Fe} = 0.1566\ \text{mol} \times 2 = 0.3132\ \text{mol}$$

Step5: Calculate mass of Fe

$$\text{Mass of Fe} = 0.3132\ \text{mol} \times 55.85\ \text{g/mol} \approx 17.5\ \text{g}$$
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Problem 8

Step1: Balance the decomposition reaction

The reaction for recovering Ag from $\text{Ag}_2\text{S}$ (simplified):

$$\text{Ag}_2\text{S} ightarrow 2\text{Ag} + \text{S}$$
Step2: Calculate molar mass of $\text{Ag}_2\text{S}$

Molar mass of $\text{Ag}_2\text{S} = 2\times107.87 + 32.07 = 247.81\ \text{g/mol}$

Step3: Find moles of $\text{Ag}_2\text{S}$

$$\text{Moles of Ag}_2\text{S} = \frac{125\ \text{g}}{247.81\ \text{g/mol}} \approx 0.504\ \text{mol}$$

Step4: Relate moles to Ag

1 mol $\text{Ag}_2\text{S}$ produces 2 mol Ag.
$$\text{Moles of Ag} = 0.504\ \text{mol} \times 2 = 1.008\ \text{mol}$$

Step5: Calculate mass of Ag

$$\text{Mass of Ag} = 1.008\ \text{mol} \times 107.87\ \text{g/mol} \approx 109\ \text{g}$$
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Answer:

  1. $\boldsymbol{39.2\ \text{g}}$
  2. $\boldsymbol{17.5\ \text{g}}$
  3. $\boldsymbol{109\ \text{g}}$