Question

What is the number of grams of cl2 formed when 0.485 mole HCl reacts with an excess of O2

Answer 1

When 0.485 mole of HCl reacts with an excess of O2, 34.43 grams of Cl2 are formed.

Step-by-step explanation:

To determine the number of grams of Cl2 formed when 0.485 mole of HCl reacts with an excess of O2, we can use the balanced chemical equation:

HCl + O2 → Cl2 + H2O

From the balanced equation, we can see that the mole ratio between HCl and Cl2 is 1:1. This means that for every mole of HCl, 1 mole of Cl2 is formed.

Given that we have 0.485 mole of HCl, we will also have 0.485 mole of Cl2 formed. To convert this to grams, we can use the molar mass of Cl2, which is 70.90 grams per mole:

0.485 mole Cl2 × 70.90 g/mol = 34.43 grams of Cl2

Answer 2

Mass of Cl₂ = 17.25 g

Step-by-step explanation:

Given data:

Moles of hydrochloric acid = 0.485 mol

Mass of chlorine gas = ?

Chemical equation:

4HCl + O₂ → 2Cl₂ + 2H₂O

Now we will compare the moles of Cl₂ with HCl .

HCl : Cl₂

4 : 2

0.485 : 2/4× 0.485 = 0.243 mol

Oxygen is present in excess that's why mass of chlorine produce depend upon the available amount of HCl.

Mass of Cl₂ :

Mass of Cl₂ = moles × molar mass

Mass of Cl₂ =0.243 mol × 71 g/mol

Mass of Cl₂ = 17.25 g