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?

Question

Naomi Aronson · Accepted Answer

#6.69# #"g P"_4#

(#"P"_4# is present in excess)

In a more lucid manner, the response is #"P"_4 (s) + 6"Cl"_2 (g) rarr 4"PCl"_3 (l)# We are required to compute the amount of excess reactant (which we must locate) that is left over after this reaction proceeds almost entirely. We must first determine how many moles of each reactant are present in order to determine the excess reactant: #"mol P"_4 = 45.13cancel("g P"_4)((1"mol P"_4)/(123.88cancel("g P"_4))) = 0.3643# #"mol P"_4# #"mol Cl"_2 = 132.0cancel("g Cl"_2)((1"mol Cl"_2)/(70.90cancel("g Cl"_2))) = 1.862# #"mol Cl"_2# We divide these values by their corresponding coefficients to determine the excess reactant; the reactant of excess is the value that comes after this. #"P"_4 = (0.3643"mol P"_4)/(1 "(coefficient)") = 0.3643# #"Cl"_2 = (1.862"mol Cl"_2)/(6 "(coefficient)") = 0.3103# Thus, the excess reactant is phosphorus. We need to use the mole value of the limiting reactant in order to find the mass of #"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 #"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.

Luna Chen · Answer

undefined To solve this problem, we need to determine which reactant is limiting and which is in excess. Then, we find out how much of the excess reactant remains after the reaction.

1. Calculate the molar mass of $ P_4 $:
   $$ P_4: 4 \times \text{Atomic mass of phosphorus} = 4 \times 30.97 \, \text{g/mol} = 123.88 \, \text{g/mol} $$

2. Calculate the molar mass of $ Cl_2 $:
   $$ Cl_2: 2 \times \text{Atomic mass of chlorine} = 2 \times 35.45 \, \text{g/mol} = 70.90 \, \text{g/mol} $$

3. Determine the number of moles for each reactant:
   $$ \text{Moles of } P_4 = \frac{45.13 \, \text{g}}{123.88 \, \text{g/mol}} = 0.364 \, \text{mol} $$
   $$ \text{Moles of } Cl_2 = \frac{132.0 \, \text{g}}{70.90 \, \text{g/mol}} = 1.860 \, \text{mol} $$

4. Calculate the mole ratio of $ P_4 $ to $ Cl_2 $ using the balanced equation:
   From the balanced equation, 1 mol of $ P_4 $ reacts with 6 mol of $ Cl_2 $.
   So, $ P_4 $ is in excess because for 0.364 mol of $ P_4 $, we would need $ 0.364 \, \text{mol} \times 6 = 2.184 \, \text{mol} $ of $ Cl_2 $ for complete reaction.

5. Calculate the mass of $ Cl_2 $ needed for the reaction:
   $$ \text{Mass of } Cl_2 = 1.860 \, \text{mol} \times 70.90 \, \text{g/mol} = 132.01 \, \text{g} $$

6. Calculate the mass of excess $ Cl_2 $ remaining:
   $$ \text{Mass of excess } Cl_2 = 132.01 \, \text{g} - 132.0 \, \text{g} = 0.01 \, \text{g} $$

Therefore, the mass of the excess reactant (Cl2) left after the reaction is 0.01 g.