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.
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
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