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

Balanced equation

PI₃ + 3 H₂O → H₃PO₃ + 3 HI

Moles of $Pl_3=\frac{150}{411.7}=0.36moles$

Limiting reagent $Pl_3$

Excess reagent $H_20$

Mass $=0.36×97.99=35.28grams$