Question #05c43

1 Answer
Feb 3, 2016

Answer:

#84.86%#

Explanation:

#2Fe(HCO_3)_3rarrFe_2(CO_3)_3+2H_2O+3CO_2#

I will assume that the #H_2O# and #CO_2# are driven off as gases.

So the mass of #H_2O# + #CO_2=1.394-0.978=0.416"g"#

We can work back and find how much iron(III) hydrogen carbonate would be needed to produce this amount of #H_2O# and #CO_2#:

2 mol #Fe(HCO_3)_3rarr# 2mol #H_2O# + 3mol #CO_2#

Converting to grams using the #M_r# values #rArr#

#(2xx238.9)"g"rarr(2xx18.01)"g"+(3xx44.00)"g"#

#:.477.8"g"rarr168.02"g"#

So this tells us that:

#168.02"g"# of #CO_2# and #H_2O# are produced from #477.8"g"# iron(III) hydrogen carbonate.

#:.1"g"# from #(477.8)/(168.02)"g"#

#:.0.416"g"# from #(477.8)/(168.02)xx0.416=1.183"g"#

This is the mass of iron(III) hydrogen carbonate.

So the % purity is:

#(1.183)/(1.394)xx100=84.86#