Answer to Question #320451 in General Chemistry for Landie

Question

Phosphorous acid (H3PO3) can be prepared from phosphorus triiodide (PI3) according to the reaction:

〖PI〗_3 (s)+ H_2 O (l)→H_3 〖PO〗_3 (aq)+HI (g)

Balance the equation. If 150 grams of 〖PI〗_3 (MM = 411.7 g/mol) is added to 250 mL of H2O (MM = 18.01 g/mol, ρ=1.00 g/mL), identify the limiting and excess reagents. How many grams of H3PO3 (MM = 81.99 g/mol) will be theoretically produced?

Answer 1

m(PI₃) = 150 g;

M(PI₃) = 411.7 g/mol;

n(PI₃) = m(PI₃)/M(PI₃) = 150/411.7 = 0.36 mol;

V(H₂O) = 250 mL;

d(H₂O) = 1.00g/mL;

m(H₂O) = d(H₂O) * V(H₂O) = 1 * 250 = 250 g;

M(H₂O) = 18.01 g/mol;

n(H₂O) = m(H₂O)/M(H₂O) = 250/18.01 = 13.88 mol;

PI₃ + 3H₂O = H₃PO₃ + 3HI;

By the chemical reaction:

n(PI₃)' = n(PI₃) = 0.36 mol;

n(H₂O)' = 1/3 * n(H₂O) = 1/3 * 13.88 = 4.63 mol.

So, PI₃ is the limiting reagent and H₂O is the excess reagent.

n(H₃PO₃) = n(PI₃) = 0.36 mol;

M(H₃PO₃) = 81.99 g/mol;

m(H₃PO₃) = n(H₃PO₃) * M(H₃PO₃) = 0.36 * 81.99 = 29.52 g.

Answer: PI₃ is the limiting reagent and H₂O is the excess reagent;

m(H₃PO₃) = 29.52 g.