The first reaction is already balanced, if you eliminate the O2 on the left side.
Your second reaction does not have Mn on both sides, and cannot be balanced.
In your third reacation, I never heard of dihydrogen monoxide. There is an unstable free radical C2O, but that cannot be what you mean. It occurs in the reaction zone of hydrocarbon flames, and C3O2 oxidation, reacting quickly with O to form extremely energetic CO*.
CuCO3 + O2 -> CuO + CO2
hydrogen peroxide + manganese dioxide -> hydrogen + oxygen
copper (II) sulfate pentahydrate + oxygen -> copper oxide + sulfur trioxide + dihydrogen monoxide
What are the balanced queations for these?
2 answers
drwls is correct about the first equation
CuCO3 -> CuO + CO2
In the second equation the manganese dioxide is a catalyst and so does not appear on either side. The products of the reaction are H2O (water but also known as dihydrogen monoxide) and oxygen
You would need to roast the copper sulfate for the third reaction but I don't think you need extra oxygen.
copper (II) sulfate pentahydrate + oxygen -> copper oxide + sulfur trioxide + dihydrogen monoxide (=water)
so you need to balance
CuSO4.5H2O -> CuO + SO3 + H2O
CuCO3 -> CuO + CO2
In the second equation the manganese dioxide is a catalyst and so does not appear on either side. The products of the reaction are H2O (water but also known as dihydrogen monoxide) and oxygen
You would need to roast the copper sulfate for the third reaction but I don't think you need extra oxygen.
copper (II) sulfate pentahydrate + oxygen -> copper oxide + sulfur trioxide + dihydrogen monoxide (=water)
so you need to balance
CuSO4.5H2O -> CuO + SO3 + H2O
0 answers