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Question4. Elemental phosphorus reacts with chlorine gas according to the equation: P4( s)+6Cl2( g)→4PCl3(l)\mathrm{P}_{4}(\mathrm{~s})+6 \mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow 4 \mathrm{PCl}_{3}(\mathrm{l})
A reaction mixture initially contains 91.38 gP491.38 \mathrm{~g} \mathrm{P}_{4} and 262.6 gCl2262.6 \mathrm{~g} \mathrm{Cl}_{2}. Once the reaction has occurred as completely as possible, what mass (in g ) of the excess reactant remains?

Studdy Solution

STEP 1

1. The chemical reaction is balanced as given.
2. The reaction goes to completion.
3. We need to find the mass of the excess reactant remaining after the reaction.

STEP 2

1. Calculate the moles of each reactant.
2. Determine the limiting reactant.
3. Calculate the amount of excess reactant remaining.

STEP 3

Calculate the moles of P4 \mathrm{P}_{4} .
The molar mass of P4 \mathrm{P}_{4} is approximately 4×30.97 g/mol=123.88 g/mol 4 \times 30.97 \, \mathrm{g/mol} = 123.88 \, \mathrm{g/mol} .
Moles of P4 \mathrm{P}_{4} = 91.38 g123.88 g/mol≈0.737 mol \frac{91.38 \, \mathrm{g}}{123.88 \, \mathrm{g/mol}} \approx 0.737 \, \mathrm{mol} .

STEP 4

Calculate the moles of Cl2 \mathrm{Cl}_{2} .
The molar mass of Cl2 \mathrm{Cl}_{2} is approximately 2×35.45 g/mol=70.90 g/mol 2 \times 35.45 \, \mathrm{g/mol} = 70.90 \, \mathrm{g/mol} .
Moles of Cl2 \mathrm{Cl}_{2} = 262.6 g70.90 g/mol≈3.703 mol \frac{262.6 \, \mathrm{g}}{70.90 \, \mathrm{g/mol}} \approx 3.703 \, \mathrm{mol} .

STEP 5

Determine the limiting reactant.
From the balanced equation, 1 mol P4 1 \, \mathrm{mol} \, \mathrm{P}_{4} reacts with 6 mol Cl2 6 \, \mathrm{mol} \, \mathrm{Cl}_{2} .
For 0.737 mol P4 0.737 \, \mathrm{mol} \, \mathrm{P}_{4} , the required moles of Cl2 \mathrm{Cl}_{2} = 0.737×6=4.422 mol 0.737 \times 6 = 4.422 \, \mathrm{mol} .
Since we only have 3.703 mol Cl2 3.703 \, \mathrm{mol} \, \mathrm{Cl}_{2} , Cl2 \mathrm{Cl}_{2} is the limiting reactant.

STEP 6

Calculate the amount of excess P4 \mathrm{P}_{4} remaining.
Since Cl2 \mathrm{Cl}_{2} is the limiting reactant, all 3.703 mol 3.703 \, \mathrm{mol} of Cl2 \mathrm{Cl}_{2} will react.
Moles of P4 \mathrm{P}_{4} that react = 3.7036≈0.617 mol \frac{3.703}{6} \approx 0.617 \, \mathrm{mol} .
Excess moles of P4 \mathrm{P}_{4} = 0.737−0.617=0.120 mol 0.737 - 0.617 = 0.120 \, \mathrm{mol} .
Mass of excess P4 \mathrm{P}_{4} = 0.120 mol×123.88 g/mol≈14.87 g 0.120 \, \mathrm{mol} \times 123.88 \, \mathrm{g/mol} \approx 14.87 \, \mathrm{g} .
The mass of the excess reactant P4 \mathrm{P}_{4} remaining is:
14.87 g \boxed{14.87 \, \mathrm{g}}

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