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Aluminum chloride, AlCl3, is an inexpensive reagent used in many industrial processes. It is made by treating scrap aluminum with chlorine according to the following equation. Please balance the equation.
__2_ Al (s) + 3___ Cl2(g) _2___ AlCl3(s)
If you start with 3.11 g of Al and 5.32 g of Cl2, which reagent is limiting? How many grams of AlCl3 can be produced? During an experiment you obtained 5.57 g of AlCl3, what was your percent yield?

Sagot :

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Answer:

83.8%

Explanation:

The balanced reaction equation is;

2Al(s) + 3Cl2(g) → 2AlCl3(s)

Now we have to obtain the limiting reactant as the reactant that produces the least amount of AlCl3

Amount of Al = 3.11g/27 g/mol = 0.115 moles

If 2 moles of Al yields 2 moles of  AlCl3

Then 0.115 moles of Al yields 0.115 moles of  AlCl3

For Cl2

Amount of Cl2 = 5.32 g/71 g/mol= 0.075 moles

If 3 moles of Cl2 yields 2 moles of  AlCl3

0.075 moles of Cl2 yields 0.075   * 2/3 = 0.05 moles of  AlCl3

Hence Cl2 is the limiting reactant

Theoretical yield of  AlCl3 = 0.05 moles of  AlCl3 * 133g/mol = 6.65 g

%yield = actual yield /theoretical yield * 100

%yield = 5.57 g/6.65 g * 100

%yield = 83.8%

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