How many grams of o2 are needed to produce 45.8 grams of Fe203 in the following
2 answers:
Explanation:
45.8 grams of Fe2O3*(1 mole/159.69 g)= moles of Fe2O3
moles of Fe2O3 *(3 moles of O2/2 moles of Fe2O3)=moles of O2
moles of O2*(32.0g of O2/1mole)= mass of O2
Answer:
Back track,
45.8 grams of Fe2O3*(1 mole/159.69 g)= moles of Fe2O3
moles of Fe2O3 *(3 moles of O2/2 moles of Fe2O3)=moles of O2
moles of O2*(32.0g of O2/1mole)= mass of O2
Explanation:
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