Question

What volume (in liters, at 703 k and 2.04 atm) of chlorine gas is required to react with 3.39 g of p?

Answer 1

The volume of chlorine required is 7.71 L.

The reaction between phosphorus and chlorine is:

2P + 5Cl₂→ 5PCl₅

Therefore, 2  moles of P requires 5 moles of chlorine to react with it.

Given mass of P =3.39 g

Molar mass of P=30.97 g/mol

No. of moles of P = given mass/ molar mass = 3.39 / 30.97 = 0.109 moles

2  moles of P requires 5 moles of chlorine

0.109  moles of P will require 0.109 x 5/2 = 0.2725 moles of chlorine

According to ideal gas equation

PV=nRT

2.04 x V = 0.2725 x 0.0821 x 703

V = 0.2725 x 0.0821 x 703 / 2.04

V = 7.71L

Learn more about ideal gas equation here:

brainly.com/question/3637553

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