If you burn Na in air, the following reactions take place:
Na + O2 => A
A + O2 => B
Note: A and B are two different substances and the chemical equations are NOT balanced.
What it is the mass percentage of oxygen in substance "A"?
(A) 25.8 %
(B) 41 %
(C) 74.2 %
(D) 59 %
(E) None of the above.
17 comments:
A could be oxide Na2O, and B superoxide NaO2..Im not sure
the balanced equation for the fist one should be 4Na+O2=>2Na2O
table of solubilities allowing and oxygen is 16/44 % of the mass of Na2O so 36.36% repeating ?
i got A ... =\
i dont think whether its balanced or not has anything to do with the mass percentage though does it ..?
Then if you think that Na2O is A you say that in
23*2+16g....are....16 g of O in
100........are....x
x=25.8%
huh ?? where'd the 100 gm come in from ?
u need to get the mase percentage
if in the molar mass u have 16 g of O then yto find out the percentage u have to find how many grams u have in 100 grms of substance:D
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you dont necessarily have to find the mass percentage from 100 gm of a substance .. you can just use the molar mass ... and im 16
i dont understand that pls find the mass percentage for Na in NaOH so i can understand how u do that:)
molar mass of NaOH = 23+16+1 = 40
mass of Na = 23
mass % of Na in NaOH = 23/40 = 57.7%
this is how i've always done mass percentages. i think if u find it from 100g it just makes the "number usage" easier. like 23g of Na in 100g of a substance makes it 23% .. so i guess its less calculations or something ... but i still like the other way better. its faster
edit: 57.5%*
Hello guys.
The correct answer is A.
the percentage is 25.8%.
This are the two equations:
4Na + O2 => 2Na2O
2Na2O + O2 => 2Na2O2 (sodium peroxide).
hahaha:)) its the same thing
when u make 23/40=0.575, and then u multiply it by 100
but i make
in 40.........23
in 1003........x
its just the same thing:))
its 23*100/40=(23/40)*100
OK I dont know the mass of sodium :\ but what was the purpose of the second equation?
Charlie unwanting to log in
In some cases Na can form directly Na2O2...so the second ecuation is to know that A its not the peroxide:D
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