A 26.6 g sample of impure potassium nitrate (KNO3) was heated to complete decomposition according to the equation
2 KNO3(s) → 2 KNO2(s) + O2(g) .
After the reaction was complete, the solid residue (consisting of KNO2 and the original impurities) had a mass of 23.1 g. Assuming that only the potassium nitrate had decomposed, what was the percent KNO3 in the original sample?
Answer in units of %.
I did this: (2 *(26.6 - 23.1)*101.103*100% ) / (32.00)(26.6)
which got 84%, but it says I'm wrong. Please help
I don't understand this stoich problem ;-;
Moderators: Xen, expert, ChenBeier
-
- Newbie
- Posts: 4
- Joined: Sun Oct 08, 2023 6:38 pm
- ChenBeier
- Distinguished Member
- Posts: 1425
- Joined: Wed Sep 27, 2017 7:25 am
- Location: Berlin, Germany
Re: I don't understand this stoich problem ;-;
I got the same amount.
3,5 g O2/ 32 g/ mol is 0,11 mol
This correspond to the double of KNO3 means 0,22 mol
Multiplied with molar mass 101 g/mol = 22,1 g
Percentage is 22,1/ 26,5 = 0,83 => 83%
3,5 g O2/ 32 g/ mol is 0,11 mol
This correspond to the double of KNO3 means 0,22 mol
Multiplied with molar mass 101 g/mol = 22,1 g
Percentage is 22,1/ 26,5 = 0,83 => 83%
-
- Newbie
- Posts: 4
- Joined: Sun Oct 08, 2023 6:38 pm
Re: I don't understand this stoich problem ;-;
Ok, thank you. I actually found a slight mistake in my method!