Question

the percent yield of the reaction shown below is 37.0%. if 73.1 g of pcl5 (208.24 g/mol) are obtained when the reaction is carried out, how many grams of pcl3 (137.33 g/mol) reacted? report your answer to the tenths place and do not include units.

Answer 1

To determine the mass of PCl₃ that reacted given a 37.0% yield and an actual yield of 73.1 g of PCl₅, we calculate the theoretical yield and use stoichiometry to find the equivalent mass of PCl₃, resulting in 130.4 g.

Step-by-step explanation:

To calculate the mass of PCl₃ that reacted based on the percent yield of PCl₅, we first need to determine the theoretical yield of PCl₅. We know that the actual yield is 73.1 g and the percent yield is 37.0%. The formula to find the theoretical yield is:

theoretical yield = (actual yield / percent yield) × 100%

Plugging in the values we have, the theoretical yield is:

theoretical yield = (73.1 g / 37.0%) × 100% = 197.57 g

With the theoretical yield, we can now determine the number of moles of PCl₅ that would have been produced had the reaction gone to completion:

moles of PCl₅ = theoretical yield / molar mass of PCl₅ = 197.57 g / 208.24 g/mol = 0.949 mol

Assuming the reaction involves a 1:1 mole ratio between PCl₃ and PCl₅, moles of PCl₃ that reacted would be equal to moles of PCl₅:

moles of PCl₃ that reacted = 0.949 mol

Now we convert this to grams:

mass of PCl₃ that reacted = moles × molar mass of PCl₃ = 0.949 mol × 137.33 g/mol = 130.4 g