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A balanced chemical equation is fundamental in understanding reactions, as it ensures the conservation of mass. In our example reaction, lead(II) nitrate \(\mathrm{Pb(NO_3)_2}\) reacts with potassium iodate \(\mathrm{KIO_3}\). This reaction can be summarized as follows: \[\mathrm{Pb(NO_3)_2(aq) + 2KIO_3(aq) \rightarrow Pb(IO_3)_2(s) + 2KNO_3(aq)}\].
Here, the equation is balanced because there is an equal number of each type of atom on both sides of the equation. For instance, we have 1 lead (Pb) atom, 2 iodine (I) atoms, and 6 oxygen (O) atoms on each side. No atoms are lost or gained, they are just rearranged.
Understanding how to balance a chemical equation is crucial for predicting the products of a reaction.

When two or more reactants are involved in a chemical reaction, the limiting reactant determines the maximum amount of product that can be formed. It is the reactant that is completely used up first, stopping the reaction from proceeding further.
In our scenario, moles of \(\mathrm{Pb(NO_3)_2}\) and \(\mathrm{KIO_3}\) are calculated, yielding 0.0750 moles and 0.150 moles respectively. According to the chemical equation, 2 moles of \(\mathrm{KIO_3}\) are necessary for every mole of \(\mathrm{Pb(NO_3)_2}\). Thus, precisely 0.150 moles of \(\mathrm{KIO_3}\) will react with 0.0750 moles of \(\mathrm{Pb(NO_3)_2}\).
Interestingly, in this case, both reactants are in perfect stoichiometric balance.

Calculating moles is an essential step in stoichiometry to understand the relationships between different reactants and products. The number of moles can be determined using the molarity formula: \[\text{Moles} = \text{Molarity} \times \text{Volume (L)}\].
For \(\mathrm{Pb(NO_3)_2}\), with a molarity of 3.00 M and a volume of 25.0 mL (0.0250 L), the number of moles is \(3.00 \times 0.0250 = 0.0750\) mol. Similarly, for \(\mathrm{KIO_3}\), with 5.00 M and 30.0 mL (0.0300 L), it results in \(5.00 \times 0.0300 = 0.150\) mol.
This calculation provides a crucial foundation for further steps like identifying the limiting reactant and computing the mass of the precipitate.

The mass of the precipitate formed is the final goal of this reaction analysis. Once we determine the moles of the precipitate \(\mathrm{Pb(IO_3)_2}\), we move to calculate its mass. With 0.0750 moles of \(\mathrm{Pb(IO_3)_2}\) formed, we multiply this by the molar mass to get the mass: \[\text{Mass} = \text{Moles} \times \text{Molar Mass}\].
The molar mass of \(\mathrm{Pb(IO_3)_2}\) is calculated by adding the atomic masses of its components: \(\mathrm{Pb} = 207.2\, \mathrm{g/mol}, \ 2 \times \mathrm{I} = 253.8\, \mathrm{g/mol}, \ \text{and} \ 6 \times \mathrm{O} = 96\, \mathrm{g/mol}\).
This totals to 557.0 g/mol. Multiplying by the moles gives us \(0.0750 \times 557.0 = 41.78\, \mathrm{g}\). This value represents the mass of the \(\mathrm{Pb(IO_3)_2}\) precipitate.