Question

the reaction of NH3 and O2 forms NO and water. the NO can be used to convert P4 to P4O6, forming N2 in the process. the P4O6 can be treated with water to form H3PO3, which forms PH3 and H3PO4 when heated. how many grams of NH3 are needed to produce 20.44 g of Ph3?

Answer 1

The equations described are following:

$4NH_3+5O_2 = 4NO + 6H_2O$

$P_4+6NO=3N_2+P_4O_6$

$P_4O_6+6H_2O=4H_3PO_3$

$4H_3PO_3 =PH_3+3H_3PO_4$

The mole number of $PH_3$ formed is:

$n(PH_3)=\frac{m(PH_3)}{M(PH_3)}=\frac{20.44}{33.99}= 0.6 mol$

By the reaction on one mol of $PH_3$ there are 4 moles of $NH_3$, so the mass of the last can be deduced:

$n(NH_3) = 4n(PH_3)=4 \times 0.6 = 2.4 mol$

$m(NH_3)=n(NH_3) \times M(NH_3)=2.4 \times 17 = 40.8 g$