When the equation for the following reaction in basic solution is balance, what is the sum of the coefficients?
MnO2 + HO2^1-.....>MnO4^1-
A=11
B=31
C=14
D=9
Here are the half equations.
MnO2 + 4 OH^- ==> MnO4^- + 3e + 2H2O
H2O + 2e + HO2^- ==> 3 OH^-
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Multiply equation 1 by 2 and equation 2 by 3 and add them to obtain the final balanced ionic equation in basic solution. After you have done this don't forget to look for common ions on each side of the equation; probably you will find some that appear on both sides. You can cancel ions that appear on both side (of course you can cancel only enough on each side to keep things equal). For example, I can tell that you will have H2O on both sides and you will have OH^- on both sides. Of course the electrons will cancel.
D)9
D)9
To balance the equation, we need to make sure that the number of each type of atom is the same on both sides of the equation.
Let's start by balancing the atoms other than hydrogen and oxygen. In this case, we only have manganese (Mn). On the left side, we have one Mn atom in MnO2, and on the right side, we have one Mn atom in MnO4^1-. So the Mn atom is already balanced.
Next, let's balance the oxygen atoms. On the left side, we have two oxygen atoms in MnO2, and on the right side, we have four oxygen atoms in MnO4^1-. To balance the oxygens, we need to add two more hydroxide (OH^-) ions to the left side.
The balanced equation becomes:
MnO2 + 2OH^- → MnO4^1- + H2O
Now let's balance the hydrogen atoms. On the left side, we have two hydroxide ions (OH^-), which means we have two hydrogen atoms. On the right side, we have one water molecule (H2O), which also contains two hydrogen atoms. So the hydrogen atoms are already balanced.
The sum of the coefficients is calculated by adding up the coefficients of all the species in the balanced equation:
1 + 2 + 1 + 1 = 5
Therefore, the correct answer is not provided in the given options.
Are you ready for a little chemistry clowning around? Well, let's balance that equation and find the sum of the coefficients together!
To balance the equation, we start by looking at the atoms on both sides.
On the left side, we have an Mn atom, an O atom, and an H atom.
On the right side, we have an Mn atom and four O atoms.
Since we have excess O atoms on the right side, let's balance the O atoms first. We can do that by adding three HO2^- ions on the left side:
MnO2 + 3HO2^- .......> MnO4^-
Now, let's count the Mn and H atoms. On the left side, we have one Mn atom and three H atoms. On the right side, we still have one Mn atom but no H atoms.
To balance the H atoms, let's add six H2O molecules on the right side:
MnO2 + 3HO2^- .......> MnO4^- + 6H2O
Now, we have one Mn atom, three H atoms, four O atoms, and six H2O molecules on the right side.
The sum of the coefficients is 1 + 3 + 6 + 4 = 14.
So, the correct answer is C) 14.
I hope I didn't leave you in a chemical haze with all these coefficients flying around!