# Question #58423

Mar 11, 2017

First convert the given quantities of compounds to moles. Then apply the balanced equation. 50g $F e C {l}_{3}$ is 50/161 = 0.31 mole (you didn’t specify Fe(II) or Fe(III)). 50g $N {a}_{3} P {O}_{4}$ is 50/164 = 0.30 mole (Specific form also not specified).
$F e C {l}_{3}$ + $N {a}_{3} P {O}_{4}$$F e P {O}_{4}$ + $3 N a C l$
So, we have almost equal molar quantities of each reactant. The reaction equation shows that one mole of $F e C {l}_{3}$ reacts with one mole of $N {a}_{3} P {O}_{4}$. The maximum products that can be formed are one mole of $F e P {O}_{4}$ and three moles of NaCl for every one mole of $F e C {l}_{3}$ . This reaction would produce 0.30 * 3 = 0.9 moles of sodium chloride.