Asked by Kelsey
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
Calculate the percent yield of iron oxide in the experiement
Answers
Answered by
DrBob222
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 = ??
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 = ??
Answered by
DANGU
quation 2 above
mole=GM/MM
=23.4/84
=0.278571
=0.21mol.
mole=GM/MM
=23.4/84
=0.278571
=0.21mol.
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