Problem 3: Consider the following reaction: 3NH_(4)NO_(3)+Na_(3)PO_(4)arrow (NH_(4))_(3)PO_(4)+3NaNO_(3) Given: 50 grams of ammonium nitrate and 60 grams of sodium phosphate. What is the Maximum amount of (NH_(4))_(3)PO_(4) can be made? Your answer

To determine the maximum amount of $(NH_{4})_{3}PO_{4}$ that can be made, we need to identify the limiting reactant in the given reaction.<br /><br />The balanced chemical equation for the reaction is:<br />$3NH_{4}NO_{3} + Na_{3}PO_{4} \rightarrow (NH_{4})_{3}PO_{4} + 3NaNO_{3}$<br /><br />Given:<br />- 50 grams of ammonium nitrate ($NH_{4}NO_{3}$)<br />- 60 grams of sodium phosphate ($Na_{3}PO_{4}$)<br /><br />Step 1: Calculate the molar masses of the reactants and products.<br />- Molar mass of $NH_{4}NO_{3}$ = 80.04 g/mol<br />- Molar mass of $Na_{3}PO_{4}$ = 163.94 g/mol<br />- Molar mass of $(NH_{4})_{3}PO_{4}$ = 149.09 g/mol<br /><br />Step 2: Calculate the number of moles of each reactant.<br />- Moles of $NH_{4}NO_{3}$ = 50 g / 80.04 g/mol = 0.625 mol<br />- Moles of $Na_{3}PO_{4}$ = 60 g / 163.94 g/mol = 0.366 mol<br /><br />Step 3: Determine the limiting reactant.<br />- According to the balanced equation, 3 moles of $NH_{4}NO_{3}$ react with 1 mole of $Na_{3}PO_{4}$.<br />- The ratio of moles of $NH_{4}NO_{3}$ to $Na_{3}PO_{4}$ is 0.625 / 0.366 = 1.71, which is greater than the required ratio of 3:1.<br />- Therefore, $Na_{3}PO_{4}$ is the limiting reactant.<br /><br />Step 4: Calculate the maximum amount of $(NH_{4})_{3}PO_{4}$ that can be made.<br />- According to the balanced equation, 1 mole of $Na_{3}PO_{4}$1 mole of $(NH_{4})_{3}PO_{4}$.<br />- Since $Na_{3}PO_{4}$ is the limiting reactant, the maximum amount of $(NH_{4})_{3}PO_{4}$ that can be made is equal to the number of moles of $Na_{3}PO_{4}$, which is 0.366 mol.<br />- The mass of $(NH_{4})_{3}PO_{4}$ produced = 0.366 mol × 149.09 g/mol = 54.6 g<br /><br />Therefore, the maximum amount of $(NH_{4})_{3}PO_{4}$ that can be made is 54.6 grams.