# Question 4aed1

Sep 22, 2015

$\text{5.12 g}$

#### Explanation:

First and foremost, you need to write a balanced chemical equation for your reaction

$\textcolor{red}{3} {\text{NiCl"_text(2(aq]) + 2"Na"_3"PO"_text(4(aq]) -> "Ni"_3("PO"_4)_text(2(s]) darr + 6"NaCl}}_{\textrm{\left(a q\right]}}$

Notice that you have a $\textcolor{red}{3} : 1$ mole ratio between nickel(II) chloride and nickel(II) phosphate.

This means that the raction will produce one mole of the latter for every 3 moles of the former that take part in the reaction.

Since sodium phosphate is in excess, you know that all the moles of nickel(II) chloride will take part in the reaction.

This means that you will produce

0.050color(red)(cancel(color(black)("moles NiCl"_2))) * ("1 mole Ni"""_3("PO"_4)_2)/(color(red)(3)color(red)(cancel(color(black)("moles NiCl"_2)))) = "0.01667 moles"

of nickel(II) phosphate.

To get the mass of nickel(II) phosphate formed by the reaction, use the compound's molar mass

0.01667color(red)(cancel(color(black)("moles"))) * "307.32 g"/(1color(red)(cancel(color(black)("mole")))) = color(green)("5.12 g")#