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I actually produced 11.2 grams of lithium chloride. What is my percent yield? Use the theoretical amount for problem #3

User Shojaeddin
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To calculate the percent yield, we first need to find the theoretical yield, which is the amount of product that should have been produced based on the balanced chemical equation and the amount of reactants used.

From problem #3, we know that the balanced chemical equation for the reaction is:

2Li + Cl2 → 2LiCl

The molar mass of Li is 6.94 g/mol, and the molar mass of Cl2 is 70.90 g/mol.

Based on the given mass of Li used (5.00 g), we can calculate the number of moles of Li:

moles of Li = mass of Li / molar mass of Li = 5.00 g / 6.94 g/mol ≈ 0.720 mol

Since the reaction uses 2 moles of Li for every 1 mole of Cl2, we need half as many moles of Cl2:

moles of Cl2 = moles of Li / 2 = 0.720 mol / 2 ≈ 0.360 mol

Now we can use the moles of Cl2 to calculate the theoretical yield of LiCl:

theoretical yield = moles of Cl2 * formula weight of LiCl

where formula weight of LiCl = atomic weight of Li + atomic weight of Cl = 6.94 g/mol + 35.45 g/mol = 42.39 g/mol

theoretical yield = 0.360 mol * 42.39 g/mol = 15.26 g

The theoretical yield of LiCl is 15.26 g.

To find the percent yield, we use the formula:

percent yield = (actual yield / theoretical yield) * 100%

Substituting the given values:

percent yield = (11.2 g / 15.26 g) * 100%

percent yield ≈ 73.4%

Therefore, the percent yield of LiCl is approximately 73.4%.
User Oriaj
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