The percentage yield for the reaction pcl3 + cl2 → pcl5 is 83.2%. what mass of pcl5 is expected from the reaction of 73.7 g pcl3 with excess chlorine?

Considering the definition of percent yield, if the percentage yield for the reaction  is 83.2%, 92.97 grams of PCl₅ is expected from the reaction of 73.7 g PCl₃ with excess chlorine.

The balanced reaction is:

PCl₃ + Cl₂ → PCl₅

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

The molar mass of the compounds is:

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

Then the following rule of three can be applied: if by reaction stoichiometry 137.35 grams of PCl₃ form 208.25 grams of PCl₅, 73.7 grams of PCl₃ form how much mass of PCl₅?

amount of mass of PCl₅= 111.74 grams

The percent yield is defined as the ratio of the actual return to the theoretical return expressed as a percentage.

The percent yield is calculated as the experimental yield divided by the theoretical yield multiplied by 100%:

where the theoretical yield is the amount of product acquired through the complete conversion of all reagents in the final product, that is, it is the maximum amount of product that could be formed from the given amounts of reagents.

In this case, you know:

percent yield= 83.2%

actual yield= ?

theorical yield= 111.74 grams

Replacing in the definition of percent yields:

Solving:

actual yield=92.97 grams

If the percentage yield for the reaction  is 83.2%, 92.97 grams of PCl₅ is expected from the reaction of 73.7 g PCl₃ with excess chlorine.

Learn more about percent yield:

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92.87 g.

Explanation:

∵ The percentage yield = (actual yield/theoretical yield)*100.

From the balanced reaction:

PCl₃ + Cl₂ → PCl₅,

It is clear that 1 mol of PCl₃ reacts with 1 mol of Cl₂ to produce 1 mol of PCl₅.

n = mass/molar mass = (73.7 g)/(137.33 g/mol) = 0.536 mol.

Using cross multiplication:

1 mol of PCl₃ produce  → 1 mol of PCl₅, from stichiometry.

∴ 0.536 mol of PCl₃ produce  → 0.536 mol of PCl₅.

∴ The mass of PCl₅ (theoretical yield) = (no. of moles) * (molar mass) = (0.536 mol)*(208.24 g/mol) = 111.62 g.

∵ The percentage yield = (actual yield/theoretical yield)*100.

The percentage yield = 83.2%, theoretical yield = 111.62 g.

∴ The actual yield of PCl₅ = (The percentage yield)(theoretical yield)/100 = (83.2%)(111.62 g)/100 = 92.87 g.

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