Consider the following reaction. 2KClO3(s) → 2KCl(s) + 3O2(g) How many grams of KCl are produced from the complete reaction of 14.7 mol of KClO3?

Question

 Consider the following reaction. 2KClO3(s) → 2KCl(s) + 3O2(g) How many grams of KCl are produced from the complete reaction of 14.7 mol of KClO3? How many grams of KCl are produced from the complete reaction of 89.7 g of KClO3?

Answer 1

The first part is easy, you already have it in moles and your equation is balanced!

 

You have 14.7 mol of KClO₃

 

You want to find how many grams of KCl is produced... 

Look at the coefficients of the balanced equation and you will see that for every 2 KClO₃ that are consumed by the reaction, 2 KCl are produced, so it is a 2:2 or 1:1 mole ratio.

 

Therefore, if you have 14.7 moles of KClO₃ that completely reacts, you will form 14.7 moles of KCl

 

To see how we got that, look here:

 

14.7 mol KClO₃ * (2 moles of KCl produced / 2 moles of KClO₃ consumed)

14.7 mol KClO₃ * (2 moles of KCl produced / 2 moles of KClO₃ consumed) = 14.7 moles KCl produced

 

Now all you need to do is convert that back to grams

 

14.7 moles KCl * ((39.1+35.5) grams of KCl / 1 mole of KCl)

 

14.7 mol KCl * (74.6 grams KCl / 1 mol KCl) = 1097 grams of KCl or to three sig figs: 1.10 x 10³ grams KCl will form from 14.7 moles of KClO₃ reacted completely

 

 

The second part of your question gives you a mass of KClO₃ so you will first have to convert to moles and then repeat the process:

 

89.7 g KClO₃ * (1 mole KClO₃ / (39.1+35.5+3*16)g KClO₃)

= 89.7 g KClO₃ * (1 mole KClO₃ / 122.6 g KClO₃)

= 0.732 moles of KClO₃

 

The mole ratio is still 1:1 so it means you form 0.732 moles of KCl

 

0.732 moles KCl * ((39.1+35.5) grams of KCl / 1 mole of KCl)

0.732 mol KCl * (74.6 grams KCl / 1 mol KCl) = 54.6 grams of KCl will form from 89.7 grams of KClO₃ reacted completely