These are the other equations I've balanced. I need help with one.
AgNO3 + K2Cr2O7 --> double replacement
AgNO3 + K2Cr2O7 --> AgCr2O7 + K2NO3
Please tell me if I've balanced these right.
Br2 + NiI3 --> single replacement, element is a non-metal
Br2 + NiI3 --> I3 + NiBr2
Cl2 + Mg3N2 --> single replacement, element is a non-metal
Cl2 + Mg3N2 --> N2 + Mg3Cl2
The equation I'm stuck on is....
HCl + Mo(OH)2 --> Neutralization
(acid-base)
Could you please show me step by step how to balance this equation. Thank you for your assistance.
AgNO3 + K2Cr2O7 --> double replacement
AgNO3 + K2Cr2O7 --> AgCr2O7 + K2NO3
The formula for silver dichromate is Ag2Cr2O7 and for potassium nitrate is KNO3. Of course this needs to be balanced again.
Please tell me if I've balanced these right.
Br2 + NiI3 --> single replacement, element is a non-metal
Br2 + NiI3 --> I3 + NiBr2
Br2 + NiI3 ==> I2 + NiBr3 and this needs to be balanced again. What are you doing wrong in these first two? Your aren't paying attention to the valence of the atoms. In the first one, Ag is +1 and dichromate ion is -2. In the second one, I2 is one of those elements that are diatomic in the free state. Remember a few days ago I wrote those as H2, N2, O2, F2, Br2, Cl2, I2, and their cousins P4 and S8. Ni in NiI3 must be +3 (since I is -1) so the formula for nickel(III) bromide is NiBr3.
Cl2 + Mg3N2 --> single replacement, element is a non-metal
Cl2 + Mg3N2 --> N2 + Mg3Cl2
Once again, you are relying on the subscript to write the formula and to help balance (maybe, maybe not) but this should be
Cl2 + Mg3N2 ==> N2 + MgCl2
The equation I'm stuck on is....
HCl + Mo(OH)2 --> Neutralization
(acid-base)
Neutralization reactions produce a salt + H2O.
HCl + Mo(OH)2 ==> H2O + MoCl2
I know the formula for the salt to be MoCl2 because Cl must be -1 (since H is +1) and Mo must be +2 (since OH polyatomic ion is -1).
I look at the entire equation and see I have 2 Cl on the right so I can make that balance with a 2 for HCl.
2HCl + Mo(OH)2 ==> H2O + MoCl2
That balances the Cl. The Mo is ok. See if the 2HCl takes care of things. I have 4 H on the left (2 from 2HCl and 2 from Mo(OH)2 and that means we need a 2 for H2O on the right.
2HCl + Mo(OH)2 ==> 2H2O + MoCl2
Now we check it.
4 H on left and 4 on right.
2Cl on left and 2 on right.
1 Mo on left and 1 on right.
2 O on left and 2 on right.
zero charge on left and zero on right.
All balanced.
It was sweaty down there
Sure, I'll guide you through the steps to balance the equation HCl + Mo(OH)2.
Step 1: Write down the unbalanced equation.
HCl + Mo(OH)2
Step 2: Identify the elements and their oxidation numbers.
In HCl, hydrogen (H) has an oxidation number of +1, and chlorine (Cl) has an oxidation number of -1.
In Mo(OH)2, hydrogen (H) has an oxidation number of +1, oxygen (O) has an oxidation number of -2, and molybdenum (Mo) has an unknown oxidation number.
Step 3: Balance the complex ion in Mo(OH)2.
Since OH is a polyatomic ion with a charge of -1, and there are two OH groups in Mo(OH)2, the overall charge is -2. Therefore, the charge on molybdenum (Mo) should be +2 to balance the charges.
Step 4: Write the balanced equation using the updated oxidation numbers.
HCl + Mo(OH)2
HCl + Mo(OH)2
Step 5: Balance the hydrogen (H) atoms.
On the left side, there is one H atom in HCl, but on the right side, there are two H atoms in Mo(OH)2. To balance this, multiply HCl by 2.
2HCl + Mo(OH)2
Step 6: Balance the chlorine (Cl) atoms.
On the left side, there is one Cl atom in HCl, but on the right side, there are no Cl atoms. To balance this, multiply Mo(OH)2 by 2.
2HCl + 2Mo(OH)2
Step 7: Balance the oxygen (O) atoms.
On the right side, there are two Mo atoms, and each Mo(OH)2 contains two O atoms. Therefore, the total number of O atoms is four. On the left side, there are two O atoms in HCl. To balance this, multiply HCl by 2.
4HCl + 2Mo(OH)2
Step 8: Check the final balance.
Count the number of each type of atom on both sides of the equation:
On the left: 4 hydrogen (H), 4 chlorine (Cl), and 2 oxygen (O).
On the right: 4 hydrogen (H), 2 molybdenum (Mo), 2 oxygen (O), and 4 hydrogen (H).
The equation is now balanced:
4HCl + 2Mo(OH)2 --> 2Mo + 4H2O
Therefore, the balanced equation for the neutralization between HCl and Mo(OH)2 is 4HCl + 2Mo(OH)2 --> 2Mo + 4H2O.