Posted by Yukiko on Friday, August 29, 2008 at 2:25am.
lets make it 100 g each..
moles S=100/32=3.33
moles O2=100/32=3.22
Rebalance the equation to
2/3 S+ O2 >> 2/3 SO3
so for each 3.33 moles of O, one needs 2/3(3.33) or 2.22 moles S, and gets 2.22 moles SO3 In this case, one has excess unreacted S.
2.22 moles SO3 will have a mass of
2.22*(32+3*16)=2.2*80=176 grams
Now scale that to 1/100, and you have it. Notice that I rounded a lot, you need to do it more accurately.
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