Doing some practice for the GRE next year.
Please could you look at my working for this and tell me why I'm wrong?
$$\ce{? FeS2 + ? O2 + ? H2O = ? Fe(OH)3 + ? H2SO4}$$
My working so far is:
There must be the same number of of $\ce{FeS2}$ as $\ce{Fe(OH)3}$
There must be twice as many $\ce{H2SO4}$ as $\ce{FeS2}$
If there is one mole of $\ce{FeS2}$ for example, there must be 1 mole of $\ce{Fe(OH)3}$ and two moles of $\ce{H2SO4}$
Tot up the $\ce{H}$ on the right: $1\times 3 + 2\times 2 = 7$. $7$ on the right so $\ce{3.5 H2O}$ on the left. ($7/2 = 3.5$)
$\ce{O}$ on the right = $1\times 3 + 2\times 4 = 11$.
This mean $11$ on the left, too. I already have $\ce{3.5 H2O}$, so we must have $11-3.5 = 7.5$ $\ce{O}$'s from the $\ce{O2}$. So $\ce{3.75 O2}$ because $7.5/2 = 3.75$.
Following the rules I've written down I arrive at:
$$\ce{1.75 FeS2 + 3.75 O2 + 3.5 H2O -> 1.75 Fe(OH)3 + 3.5 H2SO4}$$
I've multiplied these by $4$ to give integers, because that's what the question asks of me, so $7, 15, 14, 7$ and $14$.
Why am I wrong?
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