Question

lan 4 15 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 4.69 g of phosphorus with 12.66 g of chlorine according to the following reaction: P4(s) + 10 C12(g) → 4 PC15 (1) Give your answer to 2 decimal places, fill in the number ONLY!

What is the theoretical yield (in grams) of phosphorus pentachloride when 3.85 g of phosphorus reacts with 25.8 g of chlorine according to the following reaction:

P_{4 (s)} + 10 Cl_{2 (g)} → 4 PCl_{5 (l)}

Use the correct number of significant figures, fill in the number ONLY!

Answer 1

1)

Answer

2.48g

Explanation

P₄(s) + 10Cl₂(g) ------> 4PCl₅(l)

stoichiometrically, 1mole of P4 reacts with two moles 10moles of Cl2

given moles of P₄ = 4.69g/ 123.88g/mol = 0.03786mol

given moles of Cl2 = 12.66g/70.90g/mol = 0.1786mol

limiting reagent is Cl2

0.1786moles of Cl2 react with 0.01786moles of P₄

Remaining moles of P4 = 0.03786mol - 0.01786mol = 0.02mol

mass of P4 reamaining = 0.02mol × 123.88g/mol = 2.48g

2)

Answer

Explanation

P₄(s) + 10Cl₂(g) -------> 4PCl₅

given moles of P4 = 3.85g/123.88g/mol = 0.03108mol

given moles of Cl2 = 25.8g/ 70.90g/mol = 0.3639mol

limiting reagent is P4

stoichiometrically , 1mole of P4 gives 4moles of PCl5

0.03108moles of P4 gives 0.12432moles of PCl5

mass of PCl5 maximum can be produced = 0.12432mol × 208.22g/mol = 4.41g

Theoretical yield = 4.41g