A mass of 5.4g of aluminum is reacted with excess copper(II) chloride as shown in the reaction in the image. How much copper can be produced by this reaction?
3 CuCl2 + Al ---> 2 AlCl3 + 3 Cu
Your equation is not balanced. The correct balanced equation is as follows:
3 CuCl2 + 2Al ---> 2 AlCl3 + 3 Cu
mols Al = grams/atomic mass = 5.4/27 = 0.2
Convert to mols Cu produced this way.
0.2 mols Al x (3 mols Cu/2 mols Al) = 0.2 x 3/2 = 0.3 mols Cu.
Then grams Cu = mols Cu x atomic mass Cu = 0.3 x 63.54 = 19 g Cu to two significant figures.
To determine how much copper can be produced in this reaction, you need to use stoichiometry. Stoichiometry is a branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction.
Here's the step-by-step process to calculate the amount of copper produced:
1. Determine the molar mass of aluminum (Al) and copper chloride (CuCl2). The molar mass of Al is 26.98 g/mol, and the molar mass of CuCl2 is 134.45 g/mol.
2. Convert the mass of aluminum to moles. Divide the given mass of aluminum (5.4 g) by the molar mass of aluminum (26.98 g/mol) to obtain the moles of aluminum.
5.4 g Al / 26.98 g/mol = 0.2 mol Al
3. Use the balanced chemical equation to determine the stoichiometric ratio between aluminum and copper. From the equation, you can see that 3 moles of copper are produced for every 1 mole of aluminum.
Therefore, the moles of copper produced would be 3 times the moles of aluminum:
0.2 mol Al * (3 mol Cu / 1 mol Al) = 0.6 mol Cu
4. Finally, convert the moles of copper to mass. Multiply the moles of copper (0.6 mol) by the molar mass of copper (63.55 g/mol) to get the mass of copper produced.
0.6 mol Cu * 63.55 g/mol = 38.13 g Cu
Therefore, 38.13 grams of copper can be produced by reacting 5.4 grams of aluminum with excess copper(II) chloride.