If we use 3.00g of CuCl2 dihydrate, what mass of aluminum is required to complete the reaction?Here is the balanced chemical equation: 3CuCl2 dihydrate + 2Al produces 3Cu + 2AlCl3 + 6H2O Full...
If we use 3.00g of CuCl2 dihydrate, what mass of aluminum is required to complete the reaction?
Here is the balanced chemical equation: 3CuCl2 dihydrate + 2Al produces 3Cu + 2AlCl3 + 6H2O
Full explanation please!!
Anytime we are dealing with the amounts of substances in a reaction, we always want to start with a balanced chemical equation.
3CuCl2*2H2O + 2Al --> 3Cu + 2AlCl3 + 6H2O
From the balanced equation, we can see the molar relationships between each substance. For every 3 moles of the copper chloride dihydrate we will need 2 moles of aluminum to form 3 moles of Cu, 2 moles of aluminum chloride and 6 moles of water. Because these relationships are in moles, we'll need to take the mass of our reactant and convert it to moles.
IN order to do that we will need the molar mass of the compound which is found by adding the atomic masses of the elements from the periodic table. For CuCl2*2H2O (we have to include the water molecules because they are "stuck" to the copper chloride. When we weigh the sample, it will include those water molecules), the molar mass is 170.5 g/mol. ONce we have converted the moles to grams, we need to look at the mole ratio between the copper compound and the moles of aluminum.
3.00 g CuCl2*2H2O (1 mol / 170.5 g) (2 mol Al / 3 mol CuCl2*2H2) (27.0 g/mol Al)
= 0.317 g Al needed to completely react with the 3.00 g of CuCl2*2H2O
The mass of CuCl2 is 3 g and we have the molar mass of CuCl2 as 134.45 g/mol
We know that
Mole = mass/molar mass
Moles of CuCl2 = 3g/(134.45 g/mol) = 0.022 mol CuCl2
now the balanced reaction is
3CuCl2 + 2Al =========> 3Cu + 2AlCl3 + 6H2O
here the molar ratio is 3 CuCl2 : 2 Al that is 3 mol of CuCl2 requries 2 mol of Al.
But we have 0.022 mol CuCl2 so
0.022 mol CuCl2 * (2 mol Al / 3 mol CuCl2)
0.015 mol of Al
so 0.015 mol Al is requried and the molar mass of Al is 27 g/mol
we have mole = mass/molar mass
mass = moles * molar mass
Mass of Al = 0.015 mol * 27 g/mol = 0.40 g Al is requried.