Chemistry Problems

Question

1.) Elemental phosphorus reacts with chlorine gas according to the equation:
P4(s)+6Cl2(g)→4PCl3(l)
A reaction mixture initially contains 45.09g P4 and 131.1g Cl2.

Once the reaction has occurred as completely as possible, what mass (in g) of the excess reactant is left?

2.) Iron(II) sulfide reacts with hydrochloric acid according to the reaction:
FeS(s)+2HCl(aq)→FeCl2(s)+H2S(g)
A reaction mixture initially contains 0.214mol FeS and 0.664mol HCl.

Once the reaction has occurred as completely as possible, what amount (in moles) of the excess reactant is left?

8/22/2014 | Joey from Newtonville, MA

Answer 1

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Chemistry Problems

1.) Elemental phosphorus reacts with chlorine gas according to the equation:
P4(s)+6Cl2(g)→4PCl3(l)
A reaction mixture initially contains 45.09g P4 and 131.1g Cl2.

Once the reaction has occurred as completely as possible, what mass (in g) of the excess reactant is left?

2.) Iron(II) sulfide reacts with hydrochloric acid according to the reaction:
FeS(s)+2HCl(aq)→FeCl2(s)+H2S(g)
A reaction mixture initially contains 0.214mol FeS and 0.664mol HCl.

Once the reaction has occurred as completely as possible, what amount (in moles) of the excess reactant is left?

8/22/2014 | Joey from Newtonville, MA

Answer 2

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Chemistry Problems

Answer 3

Question 1

45.09g of P4 ÷ 123.896 g/mol = .363 mol P4

131.1g of Cl₂ ÷ 70.91 g/mol = 1.849 mol Cl₂

The balanced equation shows you need 1 mole of P4 for every 6 moles of Cl₂

If you divide moles of Cl₂ by 6, you will get the number of moles of P4 required.

1.849 / 6 = .308 moles of P4

Given we have .363 moles of P4, we conclude P4 is in excess.

The amount left over is .363 - .308 = .055 mol = 6.81 g

You can do the second question following my example.

I will do the second question because there is a more direct and sensible way to think about and approach the question than offered by others.

This question is easier than the first because the quantities are given in moles rather than grams so we don't even need to do the conversion to moles.

So we are given 0.214 mol FeS and 0.664 mol HCl.

From the balanced equation we know for every 1 mole of FeS we require 2 moles of HCl. So for .214 mol FeS we require .214 x 2 = .428 mol HCl. The amount we have is .664 mol which is more than enough making HCl the excess reagent

Therefore .664 - .428 = .236 mol HCl left over.

Answer 4

The chemical equation when balanced shows the number of moles of each chemical by the coefficients.

This equation shows 1 mole P₄ reacts with 6 moles Cl₂.

The molar masses for each reactant are (4x30.97) or 123.88g for P₄ and (2x35.45) or 70.90g for Cl₂.

45.09 g of P₄ divided by 123.88g/mol = .3640 mol of P₄.

131.1g of Cl₂ divided by 70.90g/mol = 1.849 mol of Cl₂.

Since there is a 1:6 ratio of P₄ to Cl₂, divide 1.849 mol by 6 to get .3082 mol Cl₂ compared to .3640 mol P₄.

The difference is .0558 mol of excess P₄.

Multiply by 123.88g/mol of P₄ to get 6.913g of excess P₄.

This answer has 4 significant digits like the given data.

Answer 5

1) With our reaction formula we have to determine the limiting reactant. To do that you have to convert the initial reactants from grams to moles.

1 mol of P4 = 123.88 g so....

45.09g of P4 / 123.88 g = 0.364 moles of P4

1 mol of Cl2 = 70.9 g so.....

131.1 g of Cl2 / 70.9 = 1.8491 moles of Cl2

The next step is to look at the coefficients of the rxn formula P4(s) results in 4PCl3(l). Since the reactsant and the product both have different coefficient this is a 1:4 ratio. 0.364 moles of P4 will result in 1.456 moles of PCl3(l) if it is the limiting reactant.

If we look that the other reactant 6Cl2(g) results in 4PCl3(l). The reactant and product have different coefficient so determining how much 4PCl3(l) is produced from 1.8491 moles of Cl2 isn't as simple. To do this we step up a proportion equation

4PCl3(l) / 6Cl2(g) = x / 1.8491 moles of Cl2

4 / 6 = x / 1.8491

(4 / 6) * 1.8491 = x

x = 1.2327 moles of PCl3(l)

Since Cl2(g) produces less product than P4 (s), that means that Cl2(g) is the limiting reactant and that there will be an excess of P4 (s). Determining how much is left over is easy. Since the Cl2(g) only produced 1.2327 moles of PCl3(l) we subtact that from the amou t that P4 (s) would have produced if it was the limiting reactant.

1.456 moles of PCl3(l) - 1.2327 moles of PCl3(l) =

1.456 - 1.2327 =

0.2233 moles of PCl3(l)

and since P4 (s) and PCl3(l) have a 1:4 ratio, this answer must be divided by 4 which means 0.2233 moles of PCl3(l) = 0.0.055825 moles of P4 (s). From here all that is left is to convert moles of P4 (s) to grams of P4 (s).

1 mol of P4(s) = 123.88 g so....

0.055825 moles of P4 (s) x 123.88 g =

6.916 g of P (s) in excess <======= ANSWER

2) This problem is solved in a similar manner so I will skip a lot of the explanations

FeS(s)+2HCl(aq)→FeCl2(s)+H2S(g)

FeS(s) => FeCl2(s) (1:1 ratio)

0.214 mol FeS = 0.214 mol FeCl2

2HCl(aq) => FeCl2(s) (2:1 ratio)

0.664 mol HCl = 0.332 mol FeCl2

FeS is limiting reactant so......

0.332 mol FeCl2 - 0.214 mol FeCl2 = 0.118 mol FeCl2

2HCl(aq) => FeCl2(s) (2:1 ratio)

2HCl(aq) / FeCl2(s) = x mol HCl / 0.118 mol FeCl2

2/1 = x/ 0.118

2 * 0.118 = x

x = 0.236 mol of excess HCl <======= ANSWER

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