Question

If 58.26 g of iodine reacts with excess aluminum. If 56.11 g of aluminum iodide is actually formed in the reaction percent yield of aluminum iodide

Answer 1

The reaction is as following:

2Al + 3I₂ = 2AlI₃

The theoretical yield of the reaction is:

m(AlI₃) = m(I₂) × [2 × Mr(AlI₃)] / [3 × Mr(I₂)] = 58.26 g × 2 × 407.695 g/mol / [3 × 253.8089 g/mol] = 62.39 g

As experimental yield equals 56.11 g and theoretical yield is 62.39 g, percent yield of aluminium iodide equals:

w = (56.11 g / 62.39 g) × 100% = 89.93%

Answer: 89.93%