Equation for the reaction:
#Fe_3O_4+H_2->H_2O+Fe#
Balanced equation:
#Fe_3O_4+4H_2->4H_2O+3Fe#
Theoretically,
1 mol of #Fe_3O_4# reacts with 4 mol of #H_2# to yield 4 mol of #H_2O#
Calculating the mole of #Fe_3O_4# and #H_2#
mole#=("mass"("g"))/("molar mass"("g"/"mol")#
Molar mass of a compound is the sum of each atom molar mass in a molecule of the compund.
Obtain the molar mass of elements from a periodic table
molar mass of #Fe_3O_4=[55.845*3+15.999*4] (g/(mol))#
molar mass of #Fe_3O_4=231.531"g/"mol"#
molar mass of #H_2=1.008*2"g"/"mol" #
molar mass of #H_2=2.016"g"/"mol" #
mol of #Fe_3O_4=963.1/231.531=4.1597"mol"#
mol of #H_2=8.57/2.016=4.251"mol"#
which is the limiting reactant?
since 4 mol of #H_2# (biggest mol number) is needed for the reaction
divide the mol of #H_2# by 4 to find the respective mol of #Fe_3O_4# to react completely with all of the #H_2# available
#("mol of "H_2)/4=4.251/4"mol"=1.06275"mol"#
#H_2# is the limiting reactant, #Fe_3O_4# is said to be in excess
find the mol of water produced
4 mol of #H_2# : 4 mol of #H_2O#
#:.#4.251mol of #H_2# : 4.251 mol of #H_2O#
PERCENTAGE YIELD
PERCENTAGE YIELD#=("experimental value")/("actual value")*100%#
the value could be mass/mol/volume...
In this case I will use mol as the value.
We need experimental mol of #H_2O#
molar mass of #H_2O=[1.008*2+15.999] "g"/"mol"#
molar mass of #H_2O=18.015 "g"/"mol"#
mol of #H_2O=(52.8)/18.015=2.9309"mol"#
PERCENTAGE YIELD #=2.9309/(4.251)*100%=68.9461%#
