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The chemical reaction between lithium carbonate and aluminum trifluoride produces aluminum carbonate and lithium fluoride:
2AlF3 + 3Li2CO3 → Al2(CO3)3 + 6LiF.

You have an excess of aluminum trifluoride and 1.70 moles of lithium carbonate, which produces 2.98 moles of lithium fluoride. What is the percent yield of the reaction? Use the periodic table and this polyatomic ion resource.

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Answer:

87.6 %

Explanation:

Step 1: Write the balanced equation

2 AlF₃ + 3 Li₂CO₃ → Al₂(CO₃)₃ + 6 LiF

Step 2: Establish the appropriate molar ratio

The molar ratio of Li₂CO₃ to LiF is 3:6.

Step 3: Calculate the theoretical yield of LiF

We calculate how much LiF could be obtained from 1.70 moles of Li₂CO₃ if the percent yield were 100%.

[tex]1.70 mol Li_2CO_3 \times \frac{6molLiF}{3mol Li_2CO_3} = 3.40 molLiF[/tex]

Step 4: Calculate the percent yield of the reaction

[tex]\frac{2.98mol}{3.40mol} \times 100\% = 87.6\%[/tex]

Answer:

87.6%

Explanation:

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