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Copper is extracted from copper oxide by reacting coppor oxide with carbon. Figure 2 shows the apparatus. Figure 2 The equation for the reaction is: 2CuO(s)+C(s)arrow 2Cu(s)+CO_(2)(g) In an experiment 15.9 g of copper oxide and 1.2 g of carbon reacted. 12.7 g of copper was produced in the reaction. (f) Calculate the mass of carbon dioxide produced in this experiment. __

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Copper is extracted from copper oxide by reacting coppor oxide with carbon. Figure 2 shows the apparatus. Figure 2 The equation for the reaction is: 2CuO(s)+C(s)arrow 2Cu(s)+CO_(2)(g) In an experiment 15.9 g of copper oxide and 1.2 g of carbon reacted. 12.7 g of copper was produced in the reaction. (f) Calculate the mass of carbon dioxide produced in this experiment. __

Copper is extracted from copper oxide by reacting coppor oxide with carbon. Figure 2 shows the apparatus. Figure 2 The equation for the reaction is: 2CuO(s)+C(s)arrow 2Cu(s)+CO_(2)(g) In an experiment 15.9 g of copper oxide and 1.2 g of carbon reacted. 12.7 g of copper was produced in the reaction. (f) Calculate the mass of carbon dioxide produced in this experiment. __

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# Explanation:<br /><br />## Step 1:<br />Identify the given quantities and the chemical equation. The chemical equation for the reaction is:<br />\[2 \mathrm{CuO}(\mathrm{s}) + \mathrm{C}(\mathrm{s}) \rightarrow 2 \mathrm{Cu}(\mathrm{s}) + \mathrm{CO}_{2}(\mathrm{~g})\]<br />Given:<br />- Mass of copper oxide (\(\mathrm{CuO}\)) = \(15.9 \, \mathrm{g}\)<br />- Mass of carbon (\(\mathrm{C}\)) = \(1.2 \, \mathrm{g}\)<br />- Mass of copper (\(\mathrm{Cu}\)) produced = \(12.7 \, \mathrm{g}\)<br /><br />## Step 2:<br />Calculate the molar masses of the reactants and products:<br />- Molar mass of \(\mathrm{CuO}\) = \(63.55 \, (\mathrm{Cu}) + 16.00 \, (\mathrm{O}) = 79.55 \, \mathrm{g/mol}\)<br />- Molar mass of \(\mathrm{C}\) = \(12.01 \, \mathrm{g/mol}\)<br />- Molar mass of \(\mathrm{Cu}\) = \(63.55 \, \mathrm{g/mol}\)<br />- Molar mass of \(\mathrm{CO}_{2}\) = \(12.01 \, (\mathrm{C}) + 2 \times 16.00 \, (\mathrm{O}) = 44.01 \, \mathrm{g/mol}\)<br /><br />## Step 3:<br />Determine the moles of \(\mathrm{CuO}\) and \(\mathrm{C}\) used in the reaction:<br />\[\text{Moles of } \mathrm{CuO} = \frac{15.9 \, \mathrm{g}}{79.55 \, \mathrm{g/mol}} = 0.2 \, \mathrm{mol}\]<br />\[\text{Moles of } \mathrm{C} = \frac{1.2 \, \mathrm{g}}{12.01 \, \mathrm{g/mol}} = 0.1 \, \mathrm{mol}\]<br /><br />## Step 4:<br />Identify the limiting reactant. According to the balanced equation, 2 moles of \(\mathrm{CuO}\) react with 1 mole of \(\mathrm{C}\). Therefore:<br />\[\text{Moles of } \mathrm{CuO} \text{ required for } 0.1 \, \mathrm{mol} \, \mathrm{C} = 2 \times 0.1 \, \mathrm{mol} = 0.2 \, \mathrm{mol}\]<br />Since the moles of \(\mathrm{CuO}\) available (0.2 mol) match exactly with the requirement, \(\mathrm{CuO}\) is the limiting reactant.<br /><br />## Step 5:<br />Calculate the moles of \(\mathrm{CO}_{2}\) produced using the stoichiometry of the reaction. From the balanced equation, 2 moles of \(\mathrm{CuO}\) produce 1 mole of \(\mathrm{CO}_{2}\):<br />\[\text{Moles of } \mathrm{CO}_{2} \text{ produced} = \frac{0.2 \, \mathrm{mol} \, \mathrm{CuO}}{2} = 0.1 \, \mathrm{mol}\]<br /><br />## Step 6:<br />Calculate the mass of \(\mathrm{CO}_{2}\) produced:<br />\[\text{Mass of } \mathrm{CO}_{2} = 0.1 \, \mathrm{mol} \times 44.01 \, \mathrm{g/mol} = 4.401 \, \mathrm{g}\]<br /><br /># Answer:<br />The mass of carbon dioxide produced in this experiment is \(4.401 \, \mathrm{g}\).
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