Question

problem 7.77
for each of the following reactions 34.0 g of each reactant is present initially
2Al(s)+3Cl2(g)-->2AlCl3(s)
A) calculate the grams of product in parentheses that would be produced
(AlCl3)

Answer 1

CONCEPT: To solve theis problem we have to find the limitting reagent in the reatants. Then calculate the amount of AlCl3 formed from the limitting reagent.

2Al(s)+3Cl2(g)-->2AlCl3(s)

As per above chemical equation 2 moles Al react with 3 moles of Cl2 to form two moles of AlCl3

Therefore Mole ratio of Al and Cl = 2:3

Moles of Al taken = given mass / atomic mass = 34.0 / 26.98 = 1.26 moles

moles of Cl2 = 34 / 35.5 = 0.96 moles

Mole ratio = 1.26 : 0.96

Or Mole ratio = 2 : 1.53 ( Multiply both by 1.59)

Hence Cl2 is limitting reagent as it is present in lesser amount than required. (For 2 moles of Al 3 moles of Cl are required)

As per the reaction 3 moles of Cl2 form 2 moles of AlCl3

Or 3 moles of Cl2 = 2 moles of AlCl3

1 mole of Cl2 = 2 / 3 moles of AlCl3

0.96 moles of Cl2 will give =  (2 / 3 ) X 0.96 = 0.64 Moles

Hence 34.0 g or 0.96 moles of Cl2 will form 0.64 moles of AlCl3

Mass of AlCl3 formed = No. of moles X Molecular mass = 0.96 X 133.5 = 128.16g