Final answer:
When 13.0 grams of aluminum react with excess copper(II) chloride, 0.723 moles or 45.96 grams of copper will be produced according to stoichiometric calculations based on the balanced chemical equation.
Step-by-step explanation:
The question asks how many moles and grams of copper will be produced when 13.0 grams of aluminum (Al) react with excess copper(II) chloride (CuCl2). To answer this, we use stoichiometry and the balanced chemical equation:
2Al(s) + 3CuCl2(aq) → 2AlCl3(aq) + 3Cu(s)
First, calculate the number of moles of Al using its molar mass (26.98 g/mol):
13.0 g Al × (1 mol Al / 26.98 g Al) = 0.482 moles Al
Then use the molar ratio between Al and Cu from the equation (3 moles of Cu for every 2 moles of Al) to find the moles of Cu:
0.482 moles Al × (3 moles Cu / 2 moles Al) = 0.723 moles Cu
Convert moles of Cu to grams using the molar mass of Cu (63.55 g/mol):
0.723 moles Cu × 63.55 g/mol = 45.96 grams Cu
Therefore, 0.723 moles or 45.96 grams of copper are produced when 13.0 grams of Al are reacted with excess CuCl2.