Answer:
1.) A.) The limiting reactant is Fe.
B.) 16.17 g.
2.) 84.70 %.
Step-by-step 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 %.