Answer:
1.) A.) The limiting reactant is Fe.
B.) 16.17 g.
2.) 84.70 %.
Explanation:
For the balanced equation:
<em>2Fe(s) + O₂(g) + 2H₂O(l) → 2Fe(OH)₂(s).</em>
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.
<em>A.) Which of these reactants is the limiting reagent?
</em>
- 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.
<em>So, 0.18 mol of Fe reacts with 0.09 mol of O₂.</em>
<em>Thus, the limiting reactant is Fe.</em>
<em>The reactant in excess is O₂ (0.125 mol - 0.09 mol = 0.035 mol).</em>
<em>B.) How many grams of Fe(OH)₂ are formed?</em>
<em><u>Using cross multiplication:</u></em>
∵ 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.
<em>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?</em>
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.
<em><u>Using cross multiplication:</u></em>
∵ 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.
<em>∴ The calculated mass (no. of grams) of produced 0.036 mol of Fe(OH)₂ = no. of moles x molar mass</em> = (0.036 mol)(89.86 g/mol) = <em>3.235 g.</em>
<em>∴ The % yield = [(actual mass/calculated mass)] x 100</em> = [(2.74 g/3.235 g)] x 100 = <em>84.70 %.</em>
