Balance each of the following neutralization reactions. Express your answer as a chemical equation. Identify all of the phases in your answer.
Part A: HNO3(aq) + Ba(OH)2(s) ---> Ba(NO3)2(aq) + H2O(l)
Part B: H2SO4(aq) + Al(OH)3 ---> Al2(SO4)3(aq) + H2O(l)
See your other posts.
To balance each of the given neutralization reactions, follow these steps:
Part A: HNO3(aq) + Ba(OH)2(s) → Ba(NO3)2(aq) + H2O(l)
1. Start by counting the number of atoms for each element on both sides of the equation.
H: 1 on the left, 2 in Ba(NO3)2, 2 in H2O on the right
N: 1 on the left, 1 in Ba(NO3)2 on the right
O: 3 in HNO3 on the left, 6 in Ba(NO3)2, 1 in H2O on the right
Ba: 1 on the left, 1 in Ba(NO3)2 on the right
2. Balance the equation by adjusting the coefficients in front of the compounds.
Since H and N are already balanced, let's start with O.
Multiply H2O by 3 to balance O: HNO3(aq) + Ba(OH)2(s) → Ba(NO3)2(aq) + 3H2O(l)
Now we have:
H: 1 on the left, 6 in H2O on the right
N: 1 on the left, 1 in Ba(NO3)2 on the right
O: 3 in HNO3, 6 in Ba(NO3)2, 6 in H2O
Ba: 1 on the left, 1 in Ba(NO3)2 on the right
3. The last step is to balance Ba and H.
Multiply Ba(OH)2 by 2: HNO3(aq) + 2Ba(OH)2(s) → Ba(NO3)2(aq) + 3H2O(l)
Now we have:
H: 6 on the left, 6 in H2O on the right
N: 1 on the left, 1 in Ba(NO3)2 on the right
O: 3 in HNO3, 6 in Ba(NO3)2, 6 in H2O
Ba: 2 on the left, 2 in Ba(NO3)2 on the right
Thus, the balanced equation for Part A is:
HNO3(aq) + 2Ba(OH)2(s) → Ba(NO3)2(aq) + 3H2O(l)
Part B: H2SO4(aq) + Al(OH)3 → Al2(SO4)3(aq) + H2O(l)
1. Counting the number of atoms for each element, we have:
H: 2 in H2SO4, 2 in H2O on the right
S: 1 in H2SO4, 3 in Al2(SO4)3 on the right
O: 4 in H2SO4, 12 in Al2(SO4)3, 1 in H2O on the right
Al: 0 on the left, 2 in Al2(SO4)3 on the right
2. Start balancing with S.
Multiply H2SO4 by 3: 3H2SO4(aq) + Al(OH)3 → Al2(SO4)3(aq) + H2O(l)
Now we have:
H: 6 on the left, 2 in H2O on the right
S: 3 in H2SO4, 3 in Al2(SO4)3 on the right
O: 12 in H2SO4, 12 in Al2(SO4)3, 1 in H2O
Al: 0 on the left, 2 in Al2(SO4)3 on the right
3. Balance the H and O.
Multiply H2O by 6 and Al(OH)3 by 2: 3H2SO4(aq) + 2Al(OH)3 → Al2(SO4)3(aq) + 6H2O(l)
Now we have:
H: 18 on the left, 6 in H2O on the right
S: 3 in H2SO4, 3 in Al2(SO4)3 on the right
O: 12 in H2SO4, 12 in Al2(SO4)3, 6 in H2O
Al: 0 on the left, 2 in Al2(SO4)3 on the right
The balanced equation for Part B is:
3H2SO4(aq) + 2Al(OH)3 → Al2(SO4)3(aq) + 6H2O(l)
In both parts, (aq) represents an aqueous solution, (s) represents a solid, and (l) represents a liquid.