Question

using the balanced equation 4FE+3O2 --> 2fe2O3 calculate how many grams of Fe2O3 are produced when 42.7 grams of Fe is reacted​

Answer 1

The coefficients in the balanced reaction

4 Fe + 3 O₂   →   2 Fe₂O₃

represent the ratio of moles of each reactant to each product. That is, for every 4 moles of Fe and 3 moles of O₂, 2 moles of Fe₂O₃ are produced.

First convert the given amount of Fe into moles. Fe has a molar mass of 55.845 g/mol, so you have

42.7 g × (1/55.845 mol/g) = 0.765 mol

of Fe.

Find how many moles of Fe₂O₃ are produced given this much Fe and an abundant supply of O₂. Multiply the coefficients of the balanced reaction by 0.765/4 :

0.765 Fe + 0.573 O₂   →   0.382 Fe₂O₃

Now convert the amount of Fe₂O₃ to grams. O has a molar mass of 15.999 g/mol, so Fe₂O₃ has a total molar mass of about

2 (55.845 g/mol) + 3 (15.999 g/mol) = 159.687 g/mol

Then the reaction would produce about

0.382 mol × (158.687 g/mol) ≈ 61.0 g

of Fe₂O₃.