Write and balanced the equation which you have done (wrong) but I corrected it.(You probably just made a typo.)
4FeS + 7O2 ==> 2Fe2O3 + 4SO2
2. Convert 23.4 g FeS to moles. moles = grams/molar mass
3. Using the coefficients in the balanced equation, convert moles FeS to moles Fe2O3.
4. Now convert moles Fe2O3 to grams. g = moles x molar mass. This is the theoretical yield.
5. %yield = (actual yield/theoretical yield)*100 = ??
In an experiement 23.4 g of iron sulphide are added to excess oxygen and 16.5 grams of iron (III) oxide, Fe2O3 are produced the balanced equation for the reaction is 4FeS + 7O2 -- 2Fe2O3 + 2SO2
Calculate the percent yield of iron oxide in the experiement
2 answers
quation 2 above
mole=GM/MM
=23.4/84
=0.278571
=0.21mol.
mole=GM/MM
=23.4/84
=0.278571
=0.21mol.
1 answer