What is the theoretical yield (in grams) of Fe(OH)3 in the reaction below when 5.00 mL of 0.150 M FeCl3 are reacted with 6.00 mL of 0.200 M NaOH?
FeCl3+3 NaOH=Fe(OH)3+3 NaCl
Can someone please explain step by step what needs to be done and why?
FeCl3+3 NaOH=Fe(OH)3+3 NaCl
Can someone please explain step by step what needs to be done and why?
1 Answer
Explanation:
Start by finding the quantity available- in
Keep in mind that
The question provides multiple quantities implying that one of the two species is in excess. The theoretical yield of
#n("FeCl"_3) = c("FeCl"_3) * V("FeCl"_3)#
#color(white)(n("FeCl"_3)) = 0.150 color(white)(l) "M" xx 5.00 color(white)(l) color(purple)("ml")#
#color(white)(n("FeCl"_3)) = 0.150 color(white)(l) mol * color(red)(cancel(color(navy)("L")^color(black)(-1))) xx 10^(-3) color(white)(l) color(red)(cancel(color(navy)("L"))) * color(purple)("ml")^(-1) xx 5.00 color(white)(l) color(purple)("ml")#
#color(white)(n("FeCl"_3)) = 7.50 xx 10^(-4) color(white)(l) color(darkgreen)(mol)#
Similarly,
#n("NaOH") = c("NaOH") * V("NaOH") = 1.2 xx 10^(-3) color(white)(l) color(darkgreen)(mol)#
Comparing the ratio of the amount of the two reactants available to this reaction to that stated in the equation identifies the limiting reactant:
#(n("FeCl"_3, color(purple)("available")))/(n("NaOH", color(purple)("available"))) = color(purple)(0.625)# #(n("FeCl"_3, color(darkgreen)("from equation")))/(n("NaOH", color(darkgreen)("from equation"))) = 1/3 ~~ color(darkgreen)(0.333)#
The second ratio states that every one mole of
Hence
Molar mass of
Therefore