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blagie [28]
3 years ago
11

I need help please ​

Chemistry
1 answer:
Alex17521 [72]3 years ago
4 0

Answer: If you think about it, B. would be the most reasonable answer with the given factors.

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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 %

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