What is "if 10.5 g copper chloride react with 12.4 g aluminum?

Copper chloride (CuCl2) and aluminum (Al) can react to form copper (Cu) and aluminum chloride (AlCl3) according to the chemical equation:

3CuCl2 + 2Al -> 2AlCl3 + 3Cu

To determine the limiting reactant and the amount of product that can be formed, we need to calculate the number of moles of each reactant.

The molar mass of CuCl2 is 134.45 g/mol, and the molar mass of Al is 26.98 g/mol.

For 10.5 g of CuCl2: moles = mass / molar mass moles = 10.5 g / 134.45 g/mol moles = 0.0781 mol

For 12.4 g of Al: moles = mass / molar mass moles = 12.4 g / 26.98 g/mol moles = 0.4592 mol

From the balanced chemical equation, it is clear that 3 moles of CuCl2 react with 2 moles of Al. Therefore, the limiting reactant is the reactant that produces the least amount of product, which in this case is CuCl2.

Using the mole ratio from the balanced chemical equation, we can calculate the maximum amount of product that can be formed:

moles of CuCl2 = 0.0781 mol moles of Al = (0.0781 mol / 3 mol) * 2 mol = 0.0521 mol

This means that only 0.0521 mol of Al will react, leaving 0.026 mol of CuCl2 unreacted.

To find the actual amount of product formed, we can use the mass of the limiting reactant (Al) and the molar mass of the product (AlCl3):

mass of AlCl3 = moles * molar mass mass of AlCl3 = 0.0521 mol * 133.34 g/mol mass of AlCl3 = 6.95 g

Therefore, if 10.5 g of CuCl2 reacts with 12.4 g of Al, the maximum amount of AlCl3 that can be formed is 6.95 g.