In this question, we have to find how many grams of the product will be produced and also how much excess of the reactant will be left over after the reaction occur. The first step will be to set up the properly balanced reaction:
2 Al + 3 Cl2 -> 2 AlCl3
Now the reaction is balanced, and according to this reaction, we have that the molar ratios are:
2 moles of Al = 3 moles of Cl2
2 moles of Al = 2 moles of AlCl3
3 moles of Cl2 = 2 moles of AlCl3
We need to find the number of moles of each reactant with the given information, starting with Aluminum, we have 6.37*10^25 atoms, and using the Avogadro's number, 6.02*10^23 is equal to 1 mol, we can find the number of moles:
6.02*10^23 = 1 mol
6.37*10^25 = x moles
x = 106 moles of Al in the reaction
According to the molar ratio, if we have 106 moles of Al, we will need:
2 Al = 3 Cl2
106 Al = x Cl2
x = 159 moles of Cl2 in order for the reaction to occur
But we need to check if we actually have this value or not, and the question tells us that we have 11.7 Liters of Cl2 at STP, and at STP 1 mol of gas is equal to 22.4 Liters of volume, therefore:
22.4 = 1 mol
11.7 = x moles
x = 0.52 moles of Cl2
As you can see we have a lot of Al and a very low value of Cl2, therefore Aluminum is the excess reactant, but let's see how much excess, according to the limiting reactant, Cl2:
3 Cl2 = 2 Al
0.52 Cl2 = x Al
x = 0.35 moles of Al are needed
We only need 0.35 moles of Al in order to react with 0.52 moles of Cl2, therefore we have the following excess:
106 - 0.35 = 105.65 moles of Al in excess
Now, to determine how much product was produced, we need to use the number of moles of the limiting reactant, 0.52 moles of Cl2, and check, through the molar ratio how many moles of AlCl3 will be produced:
3 Cl2 = 2 AlCl3
0.52 Cl2 = x AlCl3
x = 0.35 moles of AlCl3 are produced
Now with the molar mass of AlCl3, 133.34g/mol, and the given number of moles, 0.35 moles, we will have:
133.34g = 1 mol
x grams = 0.35 moles
x = 46.67 grams of AlCl3 are produced
Answers:
46.67 grams of product are produced
Aluminum is the excess reactant, we have 105.65 moles as excess