To balance the chemical equation
[tex]\[ \text{Pb}\left(\text{NO}_3\right)_2(aq) + \text{Li}_2\text{SO}_4(aq) \rightarrow \text{PbSO}_4(s) + \text{LiNO}_3(aq) \][/tex]
we need to ensure that the number of each type of atom on the reactant side is equal to the number on the product side. Let's go through the atoms one by one:
1. Lead (Pb):
- On the left side, there is 1 Pb atom.
- On the right side, there is 1 Pb atom.
Lead is balanced.
2. Nitrogen (N):
- On the left side, in Pb(NO[tex]\(_3\)[/tex])[tex]\(_2\)[/tex], there are 2 NO3 groups, so there are 2 nitrogen atoms.
- On the right side, there is only 1 NO3 group in LiNO[tex]\(_3\)[/tex].
We need 2 NO3 groups on the right side to balance nitrogen.
3. Oxygen (O):
- On the left side, in Pb(NO3)2, there are 2 NO3 groups each containing 3 oxygen atoms, making a total of 6 oxygen atoms.
- On the right side, we only have 1 NO3 group, contributing 3 oxygen atoms.
To balance oxygen, we need 2 NO3 groups on the right side.
4. Lithium (Li):
- On the left side, in Li[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex], there are 2 lithium atoms.
- On the right side, there is only 1 Li atom in LiNO[tex]\(_3\)[/tex].
To balance lithium, we need 2 lithium atoms on the right side.
5. Sulfur (S) and Oxygen (O) in Sulfate (SO4):
- On the left side, we have 1 sulfur atom in Li[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex].
- On the right side, we also have 1 sulfur atom in PbSO[tex]\(_4\)[/tex].
For oxygen, we have already taken account of the oxygen atoms in the nitrate groups. The sulfate's sulfur and associated oxygen atoms are already balanced.
Thus, to balance nitrogen, oxygen, and lithium, we can place the coefficient 2 in front of LiNO3 on the product side:
[tex]\[ \text{Pb}\left(\text{NO}_3\right)_2(aq) + \text{Li}_2\text{SO}_4(aq) \rightarrow \text{PbSO}_4(s) + 2 \text{LiNO}_3(aq) \][/tex]
Therefore, the coefficient that should appear in front of [tex]\( \text{LiNO}_3 \)[/tex] to balance the equation is:
[tex]\[ \boxed{2} \][/tex]
