Math

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
Calculate the amount of excess P4 \mathrm{P}_{4} remaining.
Since Cl2 \mathrm{Cl}_{2} is the limiting reactant, all 3.703mol 3.703 \, \mathrm{mol} of Cl2 \mathrm{Cl}_{2} will react.
Moles of P4 \mathrm{P}_{4} that react = 3.70360.617mol \frac{3.703}{6} \approx 0.617 \, \mathrm{mol} .
Excess moles of P4 \mathrm{P}_{4} = 0.7370.617=0.120mol 0.737 - 0.617 = 0.120 \, \mathrm{mol} .
Mass of excess P4 \mathrm{P}_{4} = 0.120mol×123.88g/mol14.87g 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:
14.87g \boxed{14.87 \, \mathrm{g}}

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