Consider the following balanced net ionic equation:
3SO32- (aq) + 2 CrO42- (aq) + 10 H+(aq) 3SO42- (aq) + 2 Cr3+ (aq) + 3H2O
How many grams of Na2CrO4 are theoretically required to completely oxidize 37.8 g of Na2SO3
I basically set it up like this:
37.8 g Na2SO3*(1mol Na2SO3/126g Na2SO3)*(2mol Na2CrO4/3molNa2SO3)*(162g Na2CrO4/1mol Na2CrO4)=32.4 g Na2CrO4
2007-10-16 02:33:28 · 8 answers · asked by Dan 2 in Science & Mathematics ➔ Chemistry
Didn't work the numbers but your logic is sound.
2007-10-16 02:37:53 · answer #1 · answered by Brian K² 6 · 0⤊ 0⤋
Mol of Na2SO3 =37.8 g/(23*2+32+3*16)
= 0.300 mol
Mol of Na2SO3 = 3/2 Mol of Na2CrO4
Mol of Na2CrO4 =2/3 *0.300
=0.200 mol
mass of Na2CrO4 = 0.200 mol *(23*2+52+4*16)g/mol
= 32.4 g
So your solution is corect
2007-10-16 03:17:59 · answer #2 · answered by Mohammad 1 · 0⤊ 0⤋
Yup! Correct!
2007-10-16 02:44:32 · answer #3 · answered by gelo 2 · 0⤊ 0⤋
yea i think you got it right... I would have done the same thing
2007-10-16 02:39:18 · answer #4 · answered by Anonymous · 0⤊ 0⤋
yeah, looks correct
2007-10-16 02:47:59 · answer #5 · answered by meeta1704 2 · 0⤊ 0⤋
Exactly right.
2007-10-16 02:38:01 · answer #6 · answered by steve_geo1 7 · 0⤊ 0⤋
omg. thats scary. i take chem. and we dont have to do ANYTHING like that! : o
2007-10-16 02:36:30 · answer #7 · answered by Blackcashews 3 · 0⤊ 2⤋
(o.O)Y
2007-10-16 02:54:19 · answer #8 · answered by Bananaman 5 · 0⤊ 0⤋