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 %
Electrons - each atom has its own unique #
Answer:
a) 210 mL of 95% ethanol
b) 75 mL of water
Explanation:
a) We can use the dilution equation to solve this problem. C₁ and V₁ are the concentration and volume of the stock solution, respectively, while C₂ and V₂ are the concentration and volume of the diluted solution.
C₁V₁ = C₂V₂
We want to find V₁, the volume of ethanol stock we need to dilute to prepare the final solution:
V₁ = (C₂V₂) / C₁
The concentrations are represented in percentages. We substitute in the known values to calculate V₁. The units cancel to leave us with units of mL.
V₁ = (C₂V₂) / C₁ = (70%)(285mL) / (95%) = 210 mL
b) The final solution volume is 285 mL and we have added 210 mL of ethanol, so the remaining volume is from the water that we add:
(285 mL - 210 mL) = 75 mL
