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Correct Significant Figures Feedback: Your answer 0.29 nol was either rounded di

ID: 546278 • Letter: C

Question

Correct Significant Figures Feedback: Your answer 0.29 nol was either rounded di different number of significant figures than required for this part if you nned this resulit for later calculation in this item, keep all the digits and round as the final step before submitin answer Part D What mass of PC, will be produced from the given masses of bom reactants? Express your answer to three significant figures and include the appropriate units Hints Value Ungts Submit My Answers Give Up Incorrect; Try Again; 4 attempts remaining Conti Fundamentals of General, Organic, and Biological Chen McMurry/Ballantine/Hoeger

Explanation / Answer

Balanced reaction for production of PCl5 is

P4 + 10 Cl2 -----> 4 PCl5

as you didn't give the mass of reactants i am taking random value you follow the same procedure but careful in finding the excess reactant and limiting reactant and make sure you use limiting reactant to do calculation

assuming

16 g of chlorine

23 g of phosphorous

Molar mass of Chlorine = 71 g/mol

Molar mass of Phosphorous = 124 g//mol

No. of moles of Chlorine = 16 g / 71 g/mol = 0.225 moles

No. of moles of phosphorous = 23 g / 124 g/mol = 0.185 moles

for each mole of P4, 10 moles of Cl2 is required

so 0.185 moles of P4 is excess

for 0.225 moles of 0.225/10 = 0.0225 moles of P4 is required

so Cl2 is limiting reactant and P4 in excess

when we 0.225 moles of Cl2 reacts

No. of moles of PCl5 produced = 4 * 0.225 /10 = 0.09 moles of PCl5

Molar mass of PCl5 = 208.24 g/mol

Mass of PCl5 = No. of moles of PCl5 * Molar mass of PCl5 = 0.09 moles * 208.24 g/mol = 18.74 g

follow the similar procedure with the values you have but find the limiting reactant with your values and follow the procedure accordingly