3NH_(4)NO_(3)+Na_(3)PO_(4)arrow (NH_(4))_(3)PO_(4)+3NaNO_(3) 1. You have 30 grams of ammonium nitrate and 50.0 grams of sodium phosphate, What is the limiting reactant? 2. What is the excess reactant? 3. What is the maximum amount of each product produced?

1. Calculate molar masses <br /> $NH_4NO_3$: $14.01\times 2 + 1.008\times 4 + 16\times 3 = 80.05\ \text{g/mol}$ <br /> $Na_3PO_4$: $22.99\times 3 + 30.97 + 16\times 4 = 163.94\ \text{g/mol}$ <br /><br />2. Convert masses to moles <br /> $n(NH_4NO_3) = \frac{30.0}{80.05} \approx 0.374\ \text{mol}$ <br /> $n(Na_3PO_4) = \frac{50.0}{163.94} \approx 0.305\ \text{mol}$ <br /><br />3. Compare mole ratio with reaction stoichiometry <br /> Reaction ratio: $3 NH_4NO_3 : 1 Na_3PO_4$ <br /> Required $NH_4NO_3$ for $0.305\ \text{mol Na_3PO_4}$: $0.305 \times 3 = 0.915\ \text{mol}$ <br /> Actual $NH_4NO_3$ available: $0.374\ \text{mol}$ → less than required → limiting reactant is $NH_4NO_3$.<br /><br />4. Determine excess reactant <br /> $Na_3PO_4$ is excess, since $NH_4NO_3$ limits reaction.<br /><br />5. Calculate product moles from limiting reactant <br /> From $3$ mol $NH_4NO_3$ → $1$ mol $(NH_4)_3PO_4$ and $3$ mol $NaNO_3$ <br /> $n((NH_4)_3PO_4) = 0.374 \times \frac{1}{3} \approx 0.1247\ \text{mol}$ <br /> $n(NaNO_3) = 0.374 \times \frac{3}{3} \approx 0.374\ \text{mol}$ <br /><br />6. Convert product moles to mass <br /> $(NH_4)_3PO_4$ molar mass: $14.01\times 3 + 1.008\times 12 + 30.97 + 16\times 4 = 149.09\ \text{g/mol}$ <br />Mass $(NH_4)_3PO_4$ = $0.1247\times 149.09 \approx 18.57\ \text{g}$ <br /> $NaNO_3$ molar mass: $22.99 + 14.01 + 16\times 3 = 84.99\ \text{g/mol}$ <br />Mass $NaNO_3$ = $0.374\times 84.99 \approx 31.79\ \text{g}$