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?
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
Balanced equations for the given reactions are as follows:
1) CuCO3 + O2 -> CuO + CO2
Balanced equation: 2CuCO3 + O2 -> 2CuO + 2CO2
2) Hydrogen peroxide + manganese dioxide -> Hydrogen + Oxygen
Balanced equation: 2H2O2 + MnO2 -> 2H2O + O2
3) Copper (II) sulfate pentahydrate + Oxygen -> Copper oxide + Sulfur trioxide + Dihydrogen monoxide
Balanced equation: 2CuSO4 * 5H2O + O2 -> 2CuO + 2SO3 + 9H2O
To balance chemical equations, we need to ensure that the number of atoms on both sides of the equation is equal. Let's balance each of the given equations step by step:
1. CuCO3 + O2 -> CuO + CO2
To balance the equation, we need an equal number of C, O, and Cu atoms on both sides. Let's start with the carbon (C) atoms:
There is one C atom on the left side and one on the right side. So, the C is already balanced.
Now let's balance the oxygen (O) atoms:
There are three oxygen atoms on the right side (1 in CuO and 2 in CO2), whereas there is only one on the left side. We can add another O2 molecule on the left side to obtain three oxygen atoms:
CuCO3 + O2 -> CuO + CO2(2 O2)
Now let's balance the copper (Cu) atoms:
There is one Cu atom on the left side and one on the right side. So, the Cu is already balanced.
The final balanced equation is:
2CuCO3 + O2 -> 2CuO + 2CO2
2. Hydrogen peroxide + manganese dioxide -> hydrogen + oxygen
To balance this equation, we need to equalize the number of H and O atoms on both sides. Let's start with the hydrogen (H) atoms:
There are two H atoms on the left side (in hydrogen peroxide), and two on the right side (in hydrogen gas). So, H is already balanced.
Now let's balance the oxygen (O) atoms:
There are two O atoms on the left side (in hydrogen peroxide) and only one on the right side (in oxygen gas). We can add another oxygen molecule on the right side to obtain two O atoms:
Hydrogen peroxide + manganese dioxide -> hydrogen + O2
The final balanced equation is:
2H2O2 + MnO2 -> 2H2O + O2
3. Copper (II) sulfate pentahydrate + oxygen -> copper oxide + sulfur trioxide + dihydrogen monoxide
Let's balance this equation by considering each element separately. Start with copper (Cu):
There is one Cu atom on the left side and one on the right side. So, Cu is already balanced.
Then let's move to sulfur (S):
There is one S atom on the left side (in copper sulfate pentahydrate) and one on the right side (in sulfur trioxide). So, S is already balanced.
Now let's balance the oxygen (O) atoms:
There are 15 O atoms on the left side (five in copper sulfate pentahydrate and 10 in dihydrogen monoxide), and only four on the right side (two in copper oxide and two in sulfur trioxide). We can add 11 more O2 molecules on the right side to obtain 15 O atoms:
Copper (II) sulfate pentahydrate + oxygen -> copper oxide + sulfur trioxide + 11O2
The final balanced equation is:
CuSO4∙5H2O + 11O2 -> CuO + SO3 + 10H2O
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*.