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1.) Rust forms when Fe, Oâ‚‚, and Hâ‚‚O react in the balanced equation below;

2Fe(s) + Oâ‚‚(g) +2Hâ‚‚O(l) --> 2Fe(OH)â‚‚(s)

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


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


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

2.) (Using the reaction listed in question 1.) If 2.00g Fe is reacted with an excess of Oâ‚‚ and Hâ‚‚0, and a total of 2.74g of Fe(OH)â‚‚ is actually obtained, what is the % yield?

Respuesta :

superman1987

Answer:

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)â‚‚.

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 Oâ‚‚ = 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 Oâ‚‚.

Thus, the limiting reactant is Fe.

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

B.) How many grams of Fe(OH)â‚‚ 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 Oâ‚‚ and Hâ‚‚0, and a total of 2.74 g of Fe(OH)â‚‚ 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 %.

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