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 %.

> b takes place in the gas phase in an isothermal tubular reactor. the feed is one mole of a per one mole of c, an inert. the entering temperature and pressure are 427°c and 10 atm, respectively. the gas constant r = 0.08206 atm · l/mol · k.