Balance each of the following neutralization reactions.
Part A
HBr(aq)+Mg(OH)2(s)-->MgBr2(aq)+H2O(l)
Part B
H3PO4(aq)+NaOH(aq)-->Na3PO4(aq)+H2O(l)
Trial and error.
HBr(aq)+Mg(OH)2(s)-->MgBr2(aq)+H2O(l)
Here is how I do these. It takes a little practice.
I look on the right and see Mg is ok but Br is Br2. So on the left I place a 2 for HBr.
2HBr(aq)+Mg(OH)2(s)-->MgBr2(aq)+H2O(l)
That takes care of everything except H and O. Since I know the 2 for HBr is right AND the 1 for Mg(OH)2 is right, I can count up H I have4 on the left and make the right equal to that. So I have 2H from 2HBr and I have 2H from (OH)2 which makes 4 H. On the right then I must place a 2 for 2H2O.
2HBr(aq)+Mg(OH)2(s)-->MgBr2(aq)+2H2O(l)
The only thing left is to check O and it SHOULD balance. On the left I have 2O atoms (from (OH)2) and on the right I have 2 O atoms (from 2H2O). Everything balances.
Part A:
To balance the neutralization reaction HBr(aq) + Mg(OH)2(s) → MgBr2(aq) + H2O(l), we need to ensure that the same number and types of atoms are present on both sides of the equation.
First, let's balance the hydrogen atoms by adding a coefficient of 2 in front of HBr:
2HBr(aq) + Mg(OH)2(s) → MgBr2(aq) + H2O(l)
Next, we can balance the bromine atoms by adding a coefficient of 2 in front of MgBr2:
2HBr(aq) + Mg(OH)2(s) → 2MgBr2(aq) + H2O(l)
Finally, the oxygen atoms are already balanced, so we don't need to make any changes.
The balanced equation is:
2HBr(aq) + Mg(OH)2(s) → 2MgBr2(aq) + H2O(l)
Part B:
The neutralization reaction H3PO4(aq) + NaOH(aq) → Na3PO4(aq) + H2O(l) can be balanced as follows:
First, let's balance the hydrogen atoms by adding a coefficient of 3 in front of H3PO4:
3H3PO4(aq) + NaOH(aq) → Na3PO4(aq) + H2O(l)
Next, we can balance the sodium atoms by adding a coefficient of 3 in front of NaOH:
3H3PO4(aq) + 3NaOH(aq) → Na3PO4(aq) + H2O(l)
Now, the oxygen atoms are not balanced. There are 4 oxygens on the left side but only 1 on the right side. To balance the oxygens, we need to add a coefficient of 4 in front of H2O:
3H3PO4(aq) + 3NaOH(aq) → Na3PO4(aq) + 4H2O(l)
The balanced equation is:
3H3PO4(aq) + 3NaOH(aq) → Na3PO4(aq) + 4H2O(l)
To balance the given neutralization reactions, follow these steps:
Part A:
The reaction given is: HBr(aq) + Mg(OH)2(s) → MgBr2(aq) + H2O(l)
Step 1: Count the number of atoms for each element on both sides of the reaction. You'll notice that both sides have one Br and one H. However, the left side has two H atoms while the right side has only one H atom. Therefore, we need to balance the hydrogen atoms first.
Step 2: Balance the hydrogen atoms by adding a coefficient to H2O on the right side. Add a 2 in front of H2O to balance the hydrogen atoms: HBr(aq) + Mg(OH)2(s) → MgBr2(aq) + 2H2O(l)
Step 3: Next, count the number of oxygen atoms. On the left side, there are two oxygen atoms from Mg(OH)2 and on the right side, there are four oxygen atoms from MgBr2 and the water molecules. To balance the oxygen, add a coefficient of 2 in front of Mg(OH)2: HBr(aq) + 2Mg(OH)2(s) → MgBr2(aq) + 2H2O(l)
Step 4: Lastly, balance the Mg and Br atoms. On the left side, there are two Br atoms from HBr, and on the right side, there are two Br atoms from MgBr2. This means the Br atoms are already balanced. Additionally, there are two Mg atoms on the right side, so we need to balance the Mg atoms by adding a coefficient of 2 in front of MgBr2: HBr(aq) + 2Mg(OH)2(s) → 2MgBr2(aq) + 2H2O(l)
Thus, the balanced equation for Part A is: HBr(aq) + 2Mg(OH)2(s) → 2MgBr2(aq) + 2H2O(l)
Part B:
The reaction given is: H3PO4(aq) + NaOH(aq) → Na3PO4(aq) + H2O(l)
Step 1: Count the number of atoms for each element on both sides of the reaction. On the left side, there are three H atoms and four O atoms, while on the right side, there are only two H atoms and one P atom.
Step 2: Begin balancing by first balancing the hydrogen atoms. To do this, add a coefficient of 3 in front of H2O: H3PO4(aq) + NaOH(aq) → Na3PO4(aq) + 3H2O(l)
Step 3: Next, balance the oxygen atoms. On the left side, there are 4 oxygen atoms from H3PO4 and 3 oxygen atoms from the water. On the right side, there are 4 oxygen atoms from Na3PO4. So the oxygen atoms are already balanced.
Step 4: Lastly, balance the sodium (Na) and phosphorus (P) atoms. On the left side, there is only one sodium atom, and on the right side, there are three sodium atoms (Na3PO4). Therefore, add a coefficient of 3 in front of NaOH: H3PO4(aq) + 3NaOH(aq) → Na3PO4(aq) + 3H2O(l)
Thus, the balanced equation for Part B is: H3PO4(aq) + 3NaOH(aq) → Na3PO4(aq) + 3H2O(l)