Conversions of Gases in Chemistry

Question

In the reaction 2KClO3(s) --> 2KCl(s) + 3O2(g)

If 50.0 L of oxygen at STP are needed, how many gram of KClO3 must be decomposed?

Answer 1

Since you received 2 answers that are different, I thought I'd weigh in to see what I get as an answer.

At STP, 1 mole of an ideal gas = 22.4 liters. Thus...

50.0 L O₂ x 1 mole/22.4 L = 2.232 moles of O₂

From the stoichiometry of the balanced equation we have...

2.232 moles O₂ x 2 moles KClO₃/3 moles O₂ = 1.488 moles of KClO₃ are needed

mass KClO₃ needed = 1.488 moles x 122.55 g/mole = 182.4 g = 182 g (to 3 significant figures)

This agrees with the answer provided by Michael F (only done by a slightly different approach)

Answer 2

Hi Nicole,

This is a great question to learn about a shortcut (instead of using the equation, PV = nRT) for Ideal Gases! I'll walk through the logic you can use to solve this problem.

The above reaction is as follows: 2KClO3(s) --> 2KCl(s) + 3O2(g)

First I should mention that a good rule to remember is that all these chemical reactions (including stoichiometry coefficients) are always in terms of "moles". Since 50.0 L of oxygen at Standard Temperature and Pressure (STP) are needed, we first must convert the "Liters" unit into "moles". You could use the equation, PV = nRT, to solve this problem (using STP conditions, which means Temp = 273 K and Pressure = 1 atm), but there is an easier way! A fundamental property of Ideal Gases is as follows: "1 mol of the gas at STP will take up a volume of 22.4 L". So to figure out how many moles of oxygen there are, we can use the following equation:

(50.0 L O2) x [(1 mol)/(22.4 L)] = 2.232 mol O2

And since the above reaction shows 3 moles of O2 are formed from 2 moles of decomposed KClO3, we can then calculate the number of moles of KClO3, which is the next step:

(2.23 mol O2) x (2 mol KClO3)/(3 mol O2) = 1.488 mol KClO3

Finally, we can find the grams of KClO3 needed by multiplying the moles of KClO3 by its molecular weight (MW = 122.55 g/mol; you can look this up on Google or figure it out from the Periodic Table!) as follows:

1.488 mol KClO3 x (122.55 g/mol) = 182.4 g KClO3

If you must use the correct number of significant figures, your answer will be 182 g KClO3.

Note: 3 significant figures are used in the answer since 3 significant figures are given in the question (50.0 L O2)

I hope this helps! :)

-Michael

Answer 3

Answer 4

Hi Nicole

You want to get to mole comparisons, so we start with 50 liters of oxygen. One mole of a gas at STP occupies 22.4 liters. So, 50 liters of O2 contains 50/22.4 = 2.84 moles.

Now, from the balanced equation, 2 moles of KClO3 produces 3 Moles of O2. So

2( KClO3/2)3(O2) = X moles KClo3/2.84 moles O2

Solve for X ----> X = 2*2.84/3 = 1.89 moles of KClO3

Now all you need to do is convert 1.89 moles to grams

Aou can do that, right?