Answer: <u>Volume</u>
Explanation:
Cos Milky said it. Milky smart.
54 would be the correct answer i beleive
The percentage yield is 72.8 %.
<em>Step 1</em>. Calculate the <em>mass of Br₂</em>
Mass of Br₂ = 20.0 mL Br₂ × (3.10 g Br₂/1 mL Br₂) = 62.00 g Br₂
<em>Step 2</em>. Calculate the <em>theoretical yield</em>
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
<em>Step 3</em>. Calculate the <em>percentage yield
</em>
% yield = (actual yield/theoretical yield) × 100 % = (50.3 g/69.05 g) × 100 %
= 72.8 %
I believe the answer would be 16.8
