Question

B. If the expressed mass of AICI3, to be formed is 3.5 g, what should be the initial mass of the pure aluminum metal?

C. If the actual mass of AlCl3 produced is 3.1 g, what is the percentage yield of the reaction?

pls asap

Answer 1

B. 2Al + 3Cl2 = 2AlCl3

2AlCl3 = 2Al

2 × 133g AlCl3 = 2 ×27 g Al

1g AlCl3 = 2×27/(2×133) g Al

3.5g AlCl3 = 27 × 3.5 g/ 133 Al

= 0.71g Al

C. % yield = actual yield / theoretical yield × 100%

= 3.1/3.5 ×100%

= 88.57%