Asked by amber
Oxalic acid (H2C2O4) can be used to remove rust according to the following balanced equation:
Fe2O3 + 6H2C2O4 --> 2Fe(C2O4)33- +3H2O + 6H+
Calculate the number of grams of rust that can be removed by 5.00 x 102 ml of a 0.122M solution of oxalic acid.
Fe2O3 + 6H2C2O4 --> 2Fe(C2O4)33- +3H2O + 6H+
Calculate the number of grams of rust that can be removed by 5.00 x 102 ml of a 0.122M solution of oxalic acid.
Answers
Answered by
bobpursley
My first thought is since when is rust IronIII oxide?
Ok, how many moles of oxalic acid?
moles=.500l*.122=you do it.
Now, moles of IronIIIoxide=moleacid *6)
Then, convert moles of ironIII oxide to grams.
Ok, how many moles of oxalic acid?
moles=.500l*.122=you do it.
Now, moles of IronIIIoxide=moleacid *6)
Then, convert moles of ironIII oxide to grams.
Answered by
jon
6.6 grams of rust
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