# Question 02ae1

Nov 1, 2016

a) 7.08g b) 55.5%

#### Explanation:

First, we set up the equation for the reaction. Cl_2 + P_4 → PCl_5
Then we must balance it to get the correct molar ratios: 10Cl_2 + P_4 → 4PCl_5#

Next, we calculate the molar quantities in the actual reaction.
6.03g/70.9g/mole $C {l}_{2}$ = 0.085 mole
2.20g/31g/mole P = 0.071 mole

Our balanced reaction shows that we need 10 moles of $C {l}_{2}$ to react with 4 moles of P (${P}_{4}$), so chlorine is the limiting reagent. With only 0.085 moles available, we can only react with 0.034 moles of P, or about half of that available.

Because our balanced reaction shows that every 10 moles of $C {l}_{2}$ also reacts to form only 4 moles of product $P C {l}_{5}$ we also end up with only 0.034 moles.

Converting that value back into grams gives us 0.034 mol x 208.25g/mol = 7.08g $P C {l}_{5}$

If for part b) you mean only 3.93g are actually produced, the yield of the reaction is: (actual/theoretical) x 100 = % 3.93/7.08 x 100 =55.5%