# P4(s)+6Cl2(g)→4PCl3(l) A reaction mixture initially contains 45.13 g P4 and 132.0 g Cl2.Once the reaction has occurred as completely as possible, what mass (in g) of the excess reactant is left?

Jun 10, 2017

$6.69$ ${\text{g P}}_{4}$

(${\text{P}}_{4}$ is present in excess)

#### Explanation:

In a more clear form, the reaction is

${\text{P"_4 (s) + 6"Cl"_2 (g) rarr 4"PCl}}_{3} \left(l\right)$

We're asked to calculate how much of the excess reactant (which we have to find) remains after this reaction goes essentially to completion.

To find the excess reactant, we must first find the number of moles of each reactant present:

"mol P"_4 = 45.13cancel("g P"_4)((1"mol P"_4)/(123.88cancel("g P"_4))) = 0.3643 ${\text{mol P}}_{4}$

"mol Cl"_2 = 132.0cancel("g Cl"_2)((1"mol Cl"_2)/(70.90cancel("g Cl"_2))) = 1.862 ${\text{mol Cl}}_{2}$

To find the excess reactant, we divide these values by their respective coefficients; whichever number is greater after this is the reactant of excess:

"P"_4 = (0.3643"mol P"_4)/(1 "(coefficient)") = 0.3643

"Cl"_2 = (1.862"mol Cl"_2)/(6 "(coefficient)") = 0.3103

Phosphorus is thus the excess reactant.

We need to use the mole value of the limiting reactant in order to find the mass of ${\text{P}}_{4}$ used. Using dimensional analysis,

1.862"mol Cl"_2((1"mol P"_4)/(6"mol Cl"_2))((123.88"g P"_4)/(1"mol P"_4)) = color(red)(38.44 color(red)("g P"_4

Since this is how much ${\text{P}}_{4}$ is used up, we subtract this from the initial value to find how much remains:

$45.13$ "g P"_4 - color(red)(38.44 color(red)("g P"_4) = color(blue)(6.69 color(blue)("g P"_4

Thus, after the reaction goes to completion, there will be $6.69$ grams of phosphorus left over.