Question

1. Based on the appearance of your reaction in the beaker, which reagent do you think was consumed and which reagent had some left over? The aluminum was consumed, and copper was left over as seen by the reddish particles. 2. If 5.0 g of iron metal is reacted with 15.0 g of Cl2 gas, how many grams of ferric chloride will form? About 14.52 grams will form. 3. For the reaction above the amount of ferric chloride obtained in the lab was 9.15 grams. Calculate the percent yield. The percent yield would be around 63.02%. 4. What are some reasons for obtaining a percent yield of less than 100 percent? Factors such as the reactants not reacting completely, human error in the experiment, the reactants might have too large of a surface area for reaction, multiple reactions occurring within an experiment, temperature, etc.

Answer 1

Answer:

1. Al is consumed first and CuSO₄ remains left.

2. The grams of ferric chloride that forms is 14.5 g.

3. The percent yield is 63.1%

4. Trace of impurities present in the reagents, bad manipulations when preparing solutions, etc.

Explanation:

1. The reaction is:

2Al + 3CuSO₄ = Al₂(SO₄)₃ + 3Cu

The number of moles of Al is less than the number of moles of CuSO₄. Therefore, Al is the limiting reagent and CuSO₄ is the excess reagent. This means that Al is consumed first and CuSO₄ remains left.

2. The reaction is:

2Fe + 3Cl₂ = 2FeCl₃

The number of moles of Fe is:

$n_{Fe} =\frac{m_{Fe} }{MW_{Fe} } =\frac{5}{55.85} =0.0895moles$

The number of moles of Cl₂ is:

$n_{Cl2} =\frac{15}{70.9} =0.211moles$

We know that 2 moles of Fe react with 3 moles of Cl₂, thus:

2 moles Fe---------------3 moles Cl₂

0.0895 moles Fe-------X moles Cl₂

Clearing X:

$Xmoles_{Cl2} =\frac{3*0.0895}{2} =0.134moles$

It needs 0.134 moles of Cl₂ but it only has 0.211 moles, thus, Cl₂ is the excess reagent. Fe is the limiting reagent.

2 moles Fe-----------2 moles FeCl₃

0.0895 moles Fe------X moles FeCl₃

Clearing X:

$Xmoles_{FeCl3} =\frac{2*0.0895}{2} =0.0895moles$

$m_{FeCl3} =0.0895molesFeCl3*\frac{162.2gFeCl3}{1molFeCl3} =14.5g$

3. The actual yield of FeCl₃ is 9.15 g, the theoritical yield is 14.5 g, thus, ther percent yield is:

$Percent-yield=\frac{Actual-yield}{Theoritical-yield} *100=\frac{9.15}{14.5} *100=63.1$%

4. Trace of impurities present in the reagents, bad manipulations when preparing solutions, etc.