Question

In a synthesis reaction, 2.43g of aluminum is reacted with 9.22g of chlorine. Show your work using a BCA table.

A. Find the limiting reactant
B.How many grams of the product will be produced?

Answer 1

Chlorine gas is the limiting reactant

11.57 g

Step-by-step explanation:

The equation of the reaction is;

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

The limiting reactant will produce the least number of moles of AlCl3.

For Al

Number of moles reacted = 2.43 g/27 g/mol = 0.09 moles

2 moles of Al yields 2 moles of AlCl3

0.09 moles of Al yields 0.09 * 2/2 = 0.09 moles of AlCl3

For Cl2

9.22g/71g/mol = 0.13 moles

3 moles of Cl2 yields 2 moles of AlCl3

0.13 moles of Cl2 yields 0.13 * 2/3 = 0.087 moles of AlCl3

Hence Cl2 is the limiting reactant

b) Mass of product produced = 0.087 moles of AlCl3 * 133g/mol = 11.57 g