when 63.1 g of HCl react with 17.2 g of O2 how many grams of chlorine can be produced?

Answer 1

Assuming that you are referring to the same reaction as in your previous question, we have...

4HCl + O₂ ==> 2H₂O + 2Cl₂

Let's find which reactant is in limiting supply:

moles HCl present = 63.1 g x 1 mole/36.5 g = 1.73 moles

moles O₂ present = 17.2 g x 1 mole/32 g = 0.538 moles

But since it takes 4 moles HCl per mole O₂, HCl is in limiting supply and will dictate how much product can be produced.

mass of Cl₂ that can be produced = 1.73 moles HCl x 2 moles Cl₂/4 moles HCl x 71 g/mole = 61.4 g Cl₂