Answer :
With a balanced reaction, we can determine the moles of products. The balanced equation will be:
FeCl3 + 3NH4OH → Fe(OH)3 + 3NH4Cl
They says that NH4OH is in excess, so, the limiting reactant will be FeCl3 and we will do all the calculations with this reactant.
We have to calculate the moles of FeCl3, we will use the molar mass:
[tex]\begin{gathered} molFeCl_3=5.2mg\times\frac{1g}{1000mg}\times\frac{1molFeCl_3}{MolarMass,gFeCl_3} \\ molFeCl_3=5.2mg\times\frac{1g}{1,000mg}\times\frac{1molFeCl_3}{162.2gFeCl_3}=3.2\times10^{-5}molFeCl_3 \end{gathered}[/tex]Now, to calculate the moles of the products we must take into account the product/reactive ratios, for this we are guided by the coefficients that accompany the molecules.
Ratio Fe(OH)3 to FeCl3 = 1/1
Ratio NH4Cl to FeCl3 = 3/1
Moles of each product
Moles of Fe(OH)3
[tex]molFe(OH)_3=3.2\times10^{-5}molFeCl_3\times\frac{1molFe(OH)_3}{1molFeCl_3}=3.2\times10^{-5}molFe(OH)_3[/tex]Moles of NH4Cl
[tex]molNH_4Cl=3.2\times10^{-5}molFeCl_3\times\frac{3molNH_4Cl}{1molFeCl_3}=9.6\times10^{-5}molNH_4Cl[/tex]The grams of each product we will find by multiplying the moles by the molar mass. So we have.
g of Fe(OH)3
[tex]\begin{gathered} gFe(OH)_3=3.2\times10^{-5}molFe(OH)_3\times106.87g/molFe(OH)_3 \\ gFe(OH)_3=3.4\times10^{-3}g=3.4mg \end{gathered}[/tex]g of NH4Cl
[tex]\begin{gathered} gNH_4Cl=molNH_4Cl\times MolarMassNH_4Cl \\ gNH_4Cl=9.6\times10^{-5}molNH_4Cl\times53.491g/molNH_4Cl=5.1\times10^{-3}g=5.1mg \end{gathered}[/tex]