Problem 2: 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 .How much NaNO_(3) can be produced? 0.625 mol NaNO_(3) 1.10 mol NaNO_(3) 3.53 L Na_(3)PO_(4) 1.10 mol Na_(3)PO_(4) 0.625 mol Na_(3)PO_(4)

To solve this problem, we need to determine the amount of sodium nitrate (NaNO3) that can be produced from the given amounts of ammonium nitrate (NH4NO3) and sodium phosphate (Na3PO4).<br /><br />Given information:<br />- 50 grams of ammonium nitrate (NH4NO3)<br />- 60 grams of sodium phosphate (Na3PO4)<br /><br />Step 1: Calculate the molar masses of the reactants and products.<br />Molar mass of NH4NO3 = 80.05 g/mol<br />Molar mass of Na3PO4 = 163.94 g/mol<br />Molar mass of NaNO3 = 85.00 g/mol<br /><br />Step 2: Calculate the moles of the reactants.<br />Moles of NH4NO3 = 50 g / 80.05 g/mol = 0.625 mol<br />Moles of Na3PO4 = 60 g / 163.94 g/mol = 0.366 mol<br /><br />Step 3: Determine the limiting reactant.<br />The balanced chemical equation is:<br />3NH4NO3 + Na3PO4 → (NH4)3PO4 + 3NaNO3<br /><br />From the balanced equation, we can see that 3 moles of NH4NO3 react with 1 mole of Na3PO4 to produce 3 moles of NaNO3.<br /><br />Since the moles of NH4NO3 (0.625 mol) are greater than the moles of Na3PO4 (0.366 mol), Na3PO4 is the limiting reactant.<br /><br />Step 4: Calculate the amount of NaNO3 that can be produced.<br />Since Na3PO4 is the limiting reactant, the amount of NaNO3 produced will be determined by the amount of Na3PO4 available.<br /><br />Moles of NaNO3 produced = 3 × Moles of Na3PO4<br />Moles of NaNO3 produced = 3 × 0.366 mol = 1.10 mol<br /><br />Therefore, the amount of NaNO3 that can be produced is 1.10 mol NaNO3.