2 Al(s) + Fe2O3(aq) - AlO3(aq) + 2 Fe(s) You react 20.00 grams of aluminum with iron(III) oxide. How many grams of iron should you produce?
What is the percent yield if the experimental yield is 32.67 grams of iron?

78.8%

Explanation:

The equation of the reaction is;

2 Al(s) + Fe2O3(aq) > Al2O3(aq) + 2Fe(s)

Number of moles in 20g of Al= 20g/27 g/mol = 0.74 moles

From the reaction equation;

2 moles of Al yields 2 moles of Fe

0.74 moles of Al yields 0.74 moles of Fe

Hence;

Mass of Fe produced = 0.74 moles of Fe * 56 g/mol

Mass of Fe produced = 41.44 g of Fe (This is the theoretical yield of Fe)

percent yield = actual yield/ theoretical yield  * 100

actual yield = 32.67 grams of iron

percent yield = 32.67 g/41.44 g * 100

percent yield = 78.8%

2fe + 3h2so4 -> fe2(so4)3 + 3h2

it is necessary to balance equations in order to get the proper amount of reactants as well as products in order not to make a miscalculation.

you cannot change the subscripts as that would change the substance itself. for instance, diatomic elements such as h2 only have the subscript two when a gas. another example would be so4, if you were to change the subscript, the element would no longer be sulfate.

we change the coefficients instead as that only changes the number of molecules rather than the substance itself.

hope this !