QUESTION IMAGE
Question
solve the following problems.
- how many grams of oxygen can be produced from the decomposition of 100. g of $\text{kclo}_3$?
- how much iron can be recovered from 25.0 g of $\text{fe}_2\text{o}_3$?
- how much silver can be produced from 125 g of $\text{ag}_2\text{s}$?
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Problem 6
Step1: Balance the decomposition reaction
The balanced chemical equation for the decomposition of $\text{KClO}_3$ is:
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):
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):
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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- $\boldsymbol{39.2\ \text{g}}$
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