# Question 131c8

May 11, 2015

Rounded to the proper number of sig figs, the reaction will produce 54.93 g of phosphorus trichloride.

SIDE NOTE You can actually have two forms of phosphorus that can participate in this reaction, white phosphorus, ${P}_{4}$, and red phosphorus, $P$.

White phosphorus is the most common form used for this reaction, so I'll use that form. The chemical equation will be different, but the strategy is exactly the same for red phosphorus.

The balanced chemical equation for this reaction looks like this

${P}_{4} + \textcolor{red}{6} C {l}_{2} \to 4 P C {l}_{3}$

Notice that you have a $1 : \textcolor{red}{6}$ mole ratio between phosphorus and chlorine, which means that, regardless of how many moles of phosphorus react, you'll need 6 times more moles of chlorine for the reaction to take place.

So, in order to see whether or not you're dealing with a limiting reagent, calculate the number of moles available for the reaction

12.39cancel("g") * ("1 mole "P_4)/(123.9cancel("g")) = "0.1000 moles" ${P}_{4}$

and

42.54cancel("g") * ("1 mole "Cl_2)/(70.906cancel("g")) = "0.5999 moles" $C {l}_{2}$

Since 0.1000 moles of white phosphorus would have required

0.1000cancel("moles"P_4) * (color(red)(6)"moles"Cl_2)/(1cancel("mole"P_4)) = "0.6000 moles" $C {l}_{2}$

you could say that chlorine is the limiting reagent by a very, very small margin. The number of moles of white phosphorus that will react is

0.5999cancel("moles"Cl_2) * ("1 mole"P_4)/(color(red)(6)cancel("moles"Cl_2)) = "0.09998 moles" ${P}_{4}$

This is equivalent to a mass of

0.09998cancel("moles") * "123.9 g"/(1cancel("mole"P_4)) = "12.388 g"#

According to the law of mass conservation, the mass of the product must be equal to the sum of the masses of the reactants that actually participated in the reaction.

The mass of phosphorus trichloride will be

${m}_{P C {l}_{3}} = {m}_{{P}_{4}} + {m}_{C {l}_{2}}$

${m}_{{P}_{4}} = 12.388 + 42.54 = \text{54.928 g}$

The number of sig figs you have to use will really impact the answer. If you round this to four sig figs, you'll get

${m}_{P C {l}_{3}} = \textcolor{g r e e n}{\text{54.93 g}}$

Notice that this is also the amount produced when all of the white phosphorus reacts.

${m}_{P C {l}_{3}} = 12.39 + 42.54 = 54.93$