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**

## Comments