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)

(a).

7.8 gg AlAl, 24.8 gg Cl2Cl2

Express your answer using three significant figures.

Answer 1

Molar mass of Al = 26.98 g/mol


mass(Al)= 7.8 g

use:
number of mol of Al,
n = mass of Al/molar mass of Al
=(7.8 g)/(26.98 g/mol)
= 0.2891 mol

Molar mass of Cl2 = 70.9 g/mol


mass(Cl2)= 24.8 g

use:
number of mol of Cl2,
n = mass of Cl2/molar mass of Cl2
=(24.8 g)/(70.9 g/mol)
= 0.3498 mol
Balanced chemical equation is:
2 Al + 3 Cl2 ---> 2 AlCl3


2 mol of Al reacts with 3 mol of Cl2
for 0.2891 mol of Al, 0.4337 mol of Cl2 is required
But we have 0.3498 mol of Cl2

so, Cl2 is limiting reagent
we will use Cl2 in further calculation


Molar mass of AlCl3,
MM = 1*MM(Al) + 3*MM(Cl)
= 1*26.98 + 3*35.45
= 133.33 g/mol

According to balanced equation
mol of AlCl3 formed = (2/3)* moles of Cl2
= (2/3)*0.3498
= 0.2332 mol


use:
mass of AlCl3 = number of mol * molar mass
= 0.2332*1.333*10^2
= 31.09 g
Answer: 31.1 g