How would i figure what the max molar yiled of H2 is when 1.46 g Mg is added to 500 mL of 0.2M CuSO4?
the equation is
2Mg+2CuSO4+H2O--->2MgSO4+Cu2O+H2
i got 0.1M but im not sure if this is correct??
and what is the limiting reactant? i got the Mg to be it, is that right?
THANKSSSSS!!!! :)
10 points are yours if you can explainit the best! :)
2006-10-03 11:10:12 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
Well that is the equation that is given in my book..its wrong?
idk
ok i got 0.03 mol of H2! :)
does the molar yeild mean it has to be converted to molarity (moles/liter)?
thnx!
2006-10-03 11:42:30 · update #1
1.46 g Mg * mol Mg/24.31 g Mg = 0.0600 mol Mg
.5L CuSO4 solution * .2 mol/L = .1 mol CuSO4
You have .06 moles of Mg and .1 mol of CuSO4, so Magnesium is the limiting reagent. Since 2 mol of Mg produce 1 mol of H2, you can only produce .03 moles of H2.
2006-10-03 11:37:12 · answer #1 · answered by Mr. E 5 · 0⤊ 0⤋
Something is wrong in your question. When Mg reacts with CuSO4 H2 is not one of the products:
Mg + CuSO4 -> MgSO4 + Cu
Now, suppose that CuSO4 is an oxidizing agent - which is not correct - and we have the equation:
Mg + 2CuSO4 + H2O -> MgSO4 + Cu2O + H2SO4
Again we don't have hydrogen as a product.
2006-10-03 11:29:40 · answer #2 · answered by Dimos F 4 · 0⤊ 0⤋
that is correct
2006-10-03 11:17:32 · answer #3 · answered by Jagjit S 1 · 0⤊ 0⤋