Question

I need help figuring out how to solve this problem: How many moles of oxygen are formed when 58.6g of KNO3 decomposes according to the following reaction? The molar mass of KNO3 is 101.11g/mol
4KNO3(s) --> 2K2O(s) +2N2(g) +5O2(g)

Answer 1

First, you need to find the amount of \(KNO_3\) in moles. Next, use the reaction.\[4KNO_3\rightarrow 2K_2O+2N_2+5O_2\] As you can see, to produce 1 mole of \(O_2\), we need \(\frac{4}{5}\) moles of \(KNO_3\). Now you can easily calculate the amount of \(O_2\): \(n_{O_2}=\frac{4}{5}n_{KNO_3}\)

Answer 2

Okay, so I calculated the moles of KNO3, which equals 0.5796moles. But I"m confused by the 4/5. To get the KNO3 to cross out, would it need to be 5/4?

Answer 3

0.5796 moles KNO3 (5O2/4KNO) was what I thought would need to be the equation, but am I looking at this wrong?

Answer 4

Oh yeah. You are right. I did it wrong. As you cross multiply, it should be 5/4.

Answer 5

Okay great. So it would be 0.724 mols O2. If the second part of the question was "If only 12.1 grams of Oxygen are collected from the experiment, calculate the percent yield of the reaction" would you start by converting the moles of O2 to grams?

Answer 6

Yes. You can do it either ways, convert them all to moles or grams.

Answer 7

does this sound right then?
% yield = 12.1g O2/23.18g O2 x 100% = 52.2%

Answer 8

Sorry for being late. That answer sounds good to me.

Answer 9

I am gonna off now. Have a good day. :)

Answer 10

thank you for your help.