How is this question solved?!
When 21.6g of a metal X are burned in excess oxygen, 40.8g of X2O3 are formed. From this data, what is the metal X?
Write the equation.
4X + 3O2 ==> 2X2O3
Grams X used = 21.6 g
grams X2O3 formed = 40.8
How much oxygen was used?
40.8-21.6 = 19.2
Convert 19.2 g O to moles, use the equation to convert moles oxygen to moles X and use moles = g/molar mass to convert moles X to molar mass.
I would then check it and the equation to see if starting with 21.6 g X would give me 40.8 g X2O3.
Post your work if you get stuck.
I am still completely stuck. I did this and I got 102 for the molar mass. The answer should be around 26-27; the answer posted indicates that Al is the element in question. Thanks, again.
Post your work and I'll find your error. The answer IS 27.
4X + 3O2 --> 2X2O3
21.6g 40.8g
Oxygen used= 40.8g-21.6g= 19.2g
moles O2= 19.2 /MM = 19.2/32 = 0.6
3/0.6 = 2/x therefore, x = 0.4 moles
MM=40.8/0.4 = 102
Not sure what is wrong?! Thanks very much!
Down to moles oxygen = 0.6 is ok.
0.6 moles oxygen means how many moles X2O3 were formed?
0.6 moles O2 x (2 moles X2O3/3 moles O2) = 0.6*2/3 = 0.4 moles X2O3
moles = grams/molar mass
rearrange to molar mass = g/moles =
40.8/0.4 = 102 so molar mass X2O3 = 102.
So molar mass is [2*X + 3(16] = 102
2X + 48 = 102 and
2X = 54 so X = 27.
You just quit too soon. The 102 is correct but that's not the molar mass of X, it is the molar mass X2O3. So you just subtract oxygen, which is 48, from that to get X which leaves 54 and since that is two of them X is 27.
I like to check these things this way. Since we know it's Al, let's call it Al.
How much Al did we start with?
21.6/27 = 0.8 moles Al.
How much Al2O3 will it form?
0.8 moles Al x (2 moles Al2O3/4 moles Al) = 0.8 moles Al*(2/4) = 0.8* (1/2) = 0.4 moles.
0.4 moles x molar mass Al2O3 = 0.4 x 102 = 40.8 grams. And that's what was in the problem.
Good check.