Question

Need help on #129. Please help!

Answer 1

The percentage yield is 72.8 %.

Step 1. Calculate the mass of Br₂

Mass of Br₂ = 20.0 mL Br₂ × (3.10 g Br₂/1 mL Br₂) = 62.00 g Br₂

Step 2. Calculate the theoretical yield

M_r:           159.81    266.69

         2Al + 3Br₂ → 2AlBr₃

Moles of Br₂ = 62.00 g Br₂ × (1 mol Br₂/(159.81 g Br₂) = 0.3880 mol Br₂

Moles of AlBr₃ = 0.3880 mol Br₂ × (2 mol AlBr₃/(3 mol Br₂) =  0.2586 mol AlBr₃

Theor. yield of AlBr₃ = 0.2586 mol AlBr₃ × 266.99 g AlBr₃)/(1 mol AlBr₃)

= 69.05 g AlCl₃

Step 3. Calculate the percentage yield

% yield = (actual yield/theoretical yield) × 100 % = (50.3 g/69.05 g) × 100 %

= 72.8 %