1.) Rust forms when Fe, O₂, and H₂O react in the balanced equation below;

2Fe (s) + O₂ (g) + 2H₂O (l) - -> 2Fe (OH) ₂ (s)

In an experiment 10.0g of Fe is allowed to react with 4.00g of O₂

A.) Which of these reactants is the limiting reagent?

B.) How many grams of Fe (OH) ₂ are formed?

2.) (Using the reaction listed in question 1.) If 2.00g Fe is reacted with an excess of O₂ and H₂0, and a total of 2.74g of Fe (OH) ₂ is actually obtained, what is the % yield?

1.) A.) The limiting reactant is Fe.

B.) 16.17 g.

2.) 84.70 %.

Explanation:

For the balanced equation:

2Fe (s) + O₂ (g) + 2H₂O (l) → 2Fe (OH) ₂ (s).

2.0 moles of Fe reacts with 1.0 mole of oxygen and 2.0 moles of water to produce 2.0 moles of Fe (OH) ₂.

A.) Which of these reactants is the limiting reagent?

To determine the limiting reactant, we should calculate the no. of moles of reactants using the relation: n = mass/molar mass. Suppose that water is exist in excess.

no. of moles Fe = mass/atomic mass = (10.0 g) / (55.845 g/mol) = 0.179 mol ≅ 0.18 mol.

no. of moles of O₂ = mass/molar mass = (4.0 g) / (32.0 g/mol) = 0.125 mol.

Since from the balanced equation; every 2.0 moles of Fe reacts with 1.0 mole of oxygen.

So, 0.18 mol of Fe reacts with 0.09 mol of O₂.

Thus, the limiting reactant is Fe.

The reactant in excess is O₂ (0.125 mol - 0.09 mol = 0.035 mol).

B.) How many grams of Fe (OH) ₂ are formed?

Using cross multiplication:

∵ 2.0 moles of Fe produce → 2.0 moles of Fe (OH) ₂.

∴ 0.18 moles of Fe produce → 0.18 moles of Fe (OH) ₂.

∴ The mass (no. of grams) of produced 0.18 mol of Fe (OH) ₂ = no. of moles x molar mass = (0.18 mol) (89.86 g/mol) = 16.17 g.

2.) (Using the reaction listed in question 1.) If 2.00 g Fe is reacted with an excess of O₂ and H₂0, and a total of 2.74 g of Fe (OH) ₂ is actually obtained, what is the % yield?

The % yield = [ (actual mass/calculated mass) ] x 100.

The actual mass = 2.74 g.

We need to calculate the theoretical mass:

Firstly, we should calculate the no. of moles of reactants using the relation: n = mass/molar mass.

no. of moles Fe = mass/atomic mass = (2.0 g) / (55.845 g/mol) = 0.0358 mol ≅ 0.036 mol.

Using cross multiplication:

∵ 2.0 moles of Fe produce → 2.0 moles of Fe (OH) ₂.

∴ 0.036 moles of Fe produce → 0.036 moles of Fe (OH) ₂.

∴ The calculated mass (no. of grams) of produced 0.036 mol of Fe (OH) ₂ = no. of moles x molar mass = (0.036 mol) (89.86 g/mol) = 3.235 g.

∴ The % yield = [ (actual mass/calculated mass) ] x 100 = [ (2.74 g/3.235 g) ] x 100 = 84.70 %.