Balance this reaction: __ underline ( )Na_(3)PO_(4)+underline ( )KOHarrow underline ( )NaOH+underline ( )K_(3)PO_(4) 1,1,3,1 1,3,21 1,3,3,1

To balance the chemical equation, we need to ensure that the number of atoms of each element is the same on both sides of the equation. Let's start by counting the number of atoms of each element in the reactants and products:<br /><br />- Sodium (Na): 3 atoms in $\mathrm{Na}_{3}\mathrm{PO}_{4}$ and 1 atom in $\mathrm{NaOH}$<br />- Potassium (K): 1 atom in $\mathrm{KOH}$ and 3 atoms in $\mathrm{K}_{3}\mathrm{PO}_{4}$<br />- Phosphorus (P): 1 atom in $\mathrm{Na}_{3}\mathrm{PO}_{4}$ and 1 atom in $\mathrm{K}_{3}\mathrm{PO}_{4}$<br />- Oxygen (O): 4 atoms in $\mathrm{Na}_{3}\mathrm{PO}_{4}$ and 1 atom in $\mathrm{KOH}$ and $\mathrm{NaOH}$, making a total of 6 atoms in the products<br /><br />Now, let's balance the equation by adjusting the coefficients:<br /><br />- We have 3 sodium atoms on the left side, so we need 3 sodium atoms on the right side. This means we need 3 molecules of $\mathrm{NaOH}$.<br />- We have 1 potassium atom on the left side, so we need 1 potassium atom on the right side. This means we need 1 molecule of $\mathrm{KOH}$.<br />- We have 1 phosphorus atom on both sides, so no adjustment is needed for phosphorus.<br />- We have 4 oxygen atoms on the left side and 6 oxygen atoms on the right side. This means we need to adjust the coefficients of $\mathrm{Na}_{3}\mathrm{PO}_{4}$ and $\mathrm{K}_{3}\mathrm{PO}_{4}$ to balance the oxygen atoms.<br /><br />After balancing the sodium and potassium atoms, we get:<br /><br />$\mathrm{Na}_{3}\mathrm{PO}_{4} + 3\mathrm{KOH} \rightarrow 3\mathrm{NaOH} + \mathrm{K}_{3}\mathrm{PO}_{4}$<br /><br />Now, let's count the atoms again to ensure they are balanced:<br /><br />- Sodium (Na): 3 atoms on both sides<br />- Potassium (K): 3 atoms on both sides<br />- Phosphorus (P): 1 atom on both sides<br />- Oxygen (O): 4 atoms on both sides<br /><br />The equation is now balanced, and the coefficients are 1, 3, 3, and 1, respectively.