Thin Iron metal will burn in oxygen to form iron (III) oxide. Write a balanced equation for this synthesis reaction. Label the limiting and excess reactants? How much mass of the Iron (III) oxide will form from 10.0 grams of each of the reactants? What is amount of the excess reagent will remain at the completion of this reaction? Show your work.

14.25 g of Fe2O3

33.26 g of Fe2O3

0.045 moles of excess reagent.

Explanation:

The equation for the reaction is shown below, remember that noting the correct chemical reaction equation is the first step towards solving the problem.

4Fe (s) + 3O2 (g) ==> 2 Fe2O3 (s)

Since the oxygen is abundant in air, oxygen is the reactant in excess while the iron is the limiting reactant. Remember that only a thin iron metal burns according to the question.

To double check our assumption above, the reactant that gives the least mass of product is the limiting reactant.

Using iron;

If 224 g of iron yields 319.38 g of Fe2O3

10g of iron will yield 10 × 319.38/224 = 14.25 g of Fe2O3

Using oxygen

96 g of oxygen yields 319.38 g of Fe2O3

10 g of oxygen will yield 10 × 319.38/96 = 33.26 g of Fe2O3

Since iron is the limiting reagent, number of moles in 10g of iron = 10g/56gmol-1 = 0.18 moles

From the reaction equation

4 moles of iron reacts with 3 moles of oxygen

0.18 moles of iron will react with 0.18 ×3/4 = 0.135 moles of oxygen

Amount of excess reagent remaining; 0.18- 0.135 = 0.045 moles of excess reagent.

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2