Problem 2: Consider the following reaction: 3NH_(4)NO_(3)+Na_(3)PO_(4)arrow (NH4)_(3)PO_(4)+3NaNO_(3) Given: 50.0 grams of ammonium nitrate and 60.0 grams of sodium phosphate What is the percent yield if 0.611 mol NaNO_(3) were produced? 55.5% 97.8% 0.978% 0.555% 578% 17.1%

To calculate the percent yield, we need to determine the theoretical yield of sodium nitrate (NaNO3) and then compare it to the actual yield given in the problem.<br /><br />Given information:<br />- Reaction: $3NH_{4}NO_{3}+Na_{3}PO_{4}\rightarrow (NH4)_{3}PO_{4}+3NaNO_{3}$<br />- 50.0 grams of ammonium nitrate (NH4NO3)<br />- 60.0 grams of sodium phosphate (Na3PO4)<br />- Actual yield of NaNO3: 0.611 mol<br /><br />Step 1: Calculate the molar mass of sodium nitrate (NaNO3).<br />Molar mass of NaNO3 = 22.99 g/mol (Na) + 14.01 g/mol (N) + 16.00 g/mol (O) = 85.00 g/mol<br /><br />Step 2: Calculate the theoretical yield of sodium nitrate (NaNO3).<br />Theoretical yield of NaNO3 = (0.611 mol) × (85.00 g/mol) = 52.135 g<br /><br />Step 3: Calculate the percent yield.<br />Percent yield = (Actual yield / Theoretical yield) × 100%<br />Percent yield = (0.611 mol × 85.00 g/mol) / 52.135 g × 100% = 97.8%<br /><br />Therefore, the correct answer is $97.8\%$.