Question

Consider the reaction between HCl and O2: 4HCl(g)+O2(g)→2H2O(l)+2Cl2(g) When 63.1 g of HCl is allowed to react with 17.2 g of O2, 45.6 g of Cl2 is collected. You may want to reference (Pages 260 - 264) Section 8.6 while completing this problem. Part A Determine the limiting reactant for this reaction

Answer 1

The limiting reactant for this reaction is the HCl

Step-by-step explanation:

This is my reaction:

4HCl(g)+O2(g)→2H2O(l)+2Cl2(g)

Molar mass O2 = 32 g/mol

Molar mass HCl = 36,45 g/mol

Mass / Molar mass = Moles

Moles HCl : 63,1 g / 36,45 g/m = 1,73 moles

Moles O2: 17,2 g / 32g/m = 0,54 moles

4 moles of HCl react with 1 mol of O2, according to reaction, so

1,73 moles of HCl, are going to react with, how many moles of O2.

4 moles HCl ___ 1 mol O2

1,73 moles HCl ___ (1,73 . 1)/ 4 = 0,43 moles of O2

O2 is my excess reagent because I need 0,43 moles and I have 0,54 so I have moles in excess.

1 mol of O2 are going to react with 4 moles of HCl

0,54 moles of O2 are going to react with, how many moles of HCl ?

1 mol O2 ____ 4 moles HCl

0,54 mol O2 ___ (0,54 . 4)/ 1 = 2,16 moles

I need 2,16 moles to consume my moles of O2, but I only have 1,73 moles, tha's why the HCl is mi limiting reactant.