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?

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.

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.

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 %.

every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. therefore, there must be the same number of atoms of each element on each side of a chemical equation.

c) should be the answer.