Question

For the reaction shown, compute the theoretical yield of product (in grams) for each of the following initial amounts of reactants.

2Al(s)+3Cl2(g)→2AlCl3(s)

Part B

7.8 g Al, 25.2 g Cl2

Express your answer using three significant figures.

Part C

0.230 g Al, 1.10 g Cl2

Express your answer using three significant figures.

Answer 1

2 Al(s) + 3 Cl2(g) --> 2 AlCl3 (s)
molar mass of Al = 27, m.m. of Cl = 35.5, Therefore m.m of Cl2 = 71

1 mole of Cl2 reacts with 2/3 moles of Al to produce 2/3 moles AlCl3

b)
7.8 g Al =(7.7/27) moles= 0.289 moles Al
25.2 g Cl2 = (25.2/71) = 0.355 moles Cl2
0.355 moles Cl2 reacts with 0.355 * 2/3 (= 0.237) moles of Al to produce same amount (0.237 moles) of AlCl3
hence there must be some of the Al is left unreacted.

and molar mass of AlCl3 = 133.5 g
now, yield of product in grams = molar mass * no of moles = 133.5 * 0.237 = 31.63 g of AlCl3

c)
0.230 g Al=( 0.230/27) moles= 8.51*10^-3 moles Al
1.10 g Cl2 = (1.10/71)=0.0155 moles Cl2
Here, By looking at the equation, we see that all of the Al will be reacted.
From the equation we have , 2 moles Al react with 3 moles Cl2 and it produces 2 moles AlCl3
Therefore, 1 mole Al reacts with 3/2 moles of Cl2 to produce 1 mole AlCl3
8.51*10^-3 moles Al reacts with 8.51*10^-3 * 3/2 (= 0.0128) moles of Cl2 to produce 1 * 8.51*10^-3 (= 8.51*10^-3) moles of AlCl3
hence some of the Cl2 is left unreacted.
yield of product in grams = molar mass * no of moles = 133.5 * 8.51*10^-3 = 1.13 g of AlCl3