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.
I don't understand this question?
What is the maximum amount in grams of SO3 that can be produced by the reaction 1.0 g of S with 1.0 g of O2
EQUATION: 2 S(s) +3 O2(g)--->2 SO3(g)
The choices are:
A) 3.8
B) 2.5
C) 0.27
D) 1.7
Shouldn't it be 2.0, or am I forgetting to do something
Thanks!
1 answer
1 answer