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

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 = ??

quation 2 above

mole=GM/MM
=23.4/84
=0.278571
=0.21mol.

To calculate the percent yield of iron oxide in the experiment, we need to compare the actual yield of iron oxide with the theoretical yield.

First, we need to determine the theoretical yield of iron (III) oxide, Fe2O3. We do this by converting the given mass of iron sulphide, FeS, to Moles using its molar mass, and then using the balanced equation to calculate the stoichiometric ratio of FeS to Fe2O3.

The molar mass of FeS is:
Mass of Fe (55.85 g/mol) + Mass of S (32.06 g/mol) = 87.91 g/mol

Converting the mass of FeS to moles:
23.4 g FeS * (1 mol FeS / 87.91 g FeS) = 0.266 mol FeS

Using the balanced equation, the stoichiometric ratio of FeS to Fe2O3 is 4:2, meaning that 4 moles of FeS will produce 2 moles of Fe2O3.

So, 0.266 mol FeS * (2 mol Fe2O3 / 4 mol FeS) = 0.133 mol Fe2O3 (theoretical yield)

Next, we need to calculate the actual yield of Fe2O3. Given that 16.5 g of Fe2O3 were produced, we convert this mass to moles using the molar mass of Fe2O3.

The molar mass of Fe2O3 is:
(Mass of Fe (55.85 g/mol) * 2) + (Mass of O (16.00 g/mol) * 3) = 159.69 g/mol

Converting the mass of Fe2O3 to moles:
16.5 g Fe2O3 * (1 mol Fe2O3 / 159.69 g Fe2O3) = 0.103 mol Fe2O3 (actual yield)

Now we can calculate the percent yield using the formula:

Percent Yield = (Actual Yield / Theoretical Yield) * 100

Percent Yield = (0.103 mol Fe2O3 / 0.133 mol Fe2O3) * 100 = 77.4%

The percent yield of iron (III) oxide in the experiment is 77.4%.