Answer:
[tex]m_{Fe(OH)_3} = 0.891gFe(OH)_3[/tex]
Explanation:
Hello!
In this case, given the chemical reaction:
[tex]3NaOH(aq)+Fe(NO_3)_3(aq)\rightarrow 3NaNO_3(aq)+Fe(OH)_3(s)[/tex]
In such a way, given the volumes and molarities of each reactant, we can compute the moles of produced iron (III) hydroxide by each of them, via the 3:1 and 1:1 mole ratios:
[tex]n_{Fe(OH)_3}^{by\ NaOH}=0.0500L*0.500\frac{molNaOH}{L}*\frac{1molFe(OH)_3}{3molNaOH} =0.00833molFe(OH)_3\\\\n_{Fe(OH)_3}^{by\ Fe(NO_3)_3}=0.0750L*0.200\frac{molFe(NO_3)_3}{L}*\frac{1molFe(OH)_3}{1molFe(NO_3)_3} =0.0150molFe(OH)_3[/tex]
It means that the sodium hydroxide is the limiting reactant and 0.00833 moles of iron (III) hydroxide are produced; thus, the required mass is:
[tex]m_{Fe(OH)_3}=0.00833molFe(OH)_3*\frac{106.87gFe(OH)_3}{1molFe(OH)_3} \\\\m_{Fe(OH)_3} = 0.891gFe(OH)_3[/tex]
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