Answer to Question #269695 in General Chemistry for uflpiupi

Question

Phosphoric acid can be prepared from phosphorus triiodide according to the reaction:

    Pl₃ (s)  +  H₂O(l) -----------> H₃PO₃(aq) + HI(g)

If 150 g of PI₃ (MM = 411.7 g/mol) is added to 250 mL of H₂O (g/mol), d = 1.00 g/mL), Identify which will be limiting and excess reagents. ( How many grams of H₃PO₃ (MM = 97.99 g/mol) will be produced theoretically?

Answer 1

PI₃ + 3H₂O → H₃PO₃ + 3HI

n(PI) $= \frac{150}{411.7} = 0.364 \;mol$

m(H₂O) $= 1.00 \times 250 = 250 \;g$

M(H₂O) = 18.0 g/mol

n(H₂O) $= \frac{250}{18.0} = 13.88 \;mol$

According to the reaction for one mole of PI we need 3 moles of water. So, PI is a limiting reactant and water is an excess reactant.

n(H₃PO₃) = n(PI₃) = 0.364 mol

m(H₃PO₃) $= 0.364 \times 97.99 = 35.66 \;g$