Question

Question 4 1pts What is the mass of excess reactant (in grams) that remains after the limiting reagent produces the maximum amount of phosphorus pentachloride possible upon the reaction of 4.23 g of phosphorus with 15.02 g of chlorine according to the following reaction: P4(s)+ 10 Cl2( 4 PCI5 ( Give your answer to 2 decimal places, fill in the number ONLY! 1 pts Question 5 Fine wires of masnes Um burn readiy in oxvgen to 603 PM roduce imaanestum.axicie.1l aafi Maare reacted 10/28/2019

Answer 1

Number of moles of P4 = 4.23 g / 123.90 g/mol = 0.0341 mole

Number of moles of Cl2 = 15.02 g / 70.9060 g/mol = 0.212 mole

From the balanced equation we can say that

1 mole of P4 requires 10 mole of Cl2 so

0.0341 mole of P4 will require

= 0.0341 mole of P4 *(10 mole of Cl2 / 1 mole of P4)

= 0.341 mole of Cl2

But we have 0.212 mole of Cl2 which is in short so Cl2 is limiting reactant and P4 is an excess reactant

From the balanced equation we can say that

10 mole of Cl2 produces 4 mole of PCl5 so

0.212 mole of Cl2 will produce

= 0.212 mole of Cl2 *(4 mole of PCl5 / 10 mole of Cl2)

= 0.0848 mole of PCl5

mass of 1 mole of PCl5 = 208.24 g so

the mass of 0.0848 mole of PCl5 = 17.7 g

Therefore, the mass of PCl5 produced would be 17.7 g

From the balanced equation we can say that

10 mole of Cl2 requires 1 mole of P4 so

0.212 mole of Cl2 will require

= 0.212 mole of Cl2 *(1 mole of P4 / 10 mole of Cl2)

= 0.0212 mole of P4

number of moles of excess reactant remain after completion of reaction = 0.0341 - 0.0212 = 0.0129 mole

mass of 1 mole of P4 = 123.90 g so

the mass of 0.0212 mole of P4 = 2.63 g

Therefore, the mass of P4 remain after completion of reaction = 2.63 g