PLEASE HELP WITH QUESTION #3!!!!
1 answer:
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Answer:
The option closest to the percentage yield is option;
d. 81.1
Explanation:
The given chemical equation of the reaction is presented as follows;
CaO (s) + H₂O (l) → Ca(OH)₂
The mass of CaO in the experiment, m = 2.00 g
The volume of water with which the CaO was reacted = Excess volume of water
Number of moles = Mass/(Molar mass)
The mass of Ca(OH)₂ recovered, actual yield = 2.14 g
The molar mass of CaO = 56.0774 g/mol
The number of moles of CaO in the reaction, n₁ = 2.00 g/(56.0774 g/mol ≈ 0.036 moles
The molar mass of Ca(OH)₂ = 74.093 g/mol
The number of moles of Ca(OH)₂ in the reaction, n₂ = 2.14 g/(74.093 g/mol) ≈ 0.029 moles
From the given chemical reaction, one mole of CaO reacts with one mole of H₂O to produce one mole of Ca(OH)₂
Therefore, 0.036 moles of CaO will produce 0.036 moles of Ca(OH)₂
Mass = Number of moles × Molar mass
The mass of 0.036 moles of Ca(OH)₂ ≈ 0.036 moles × 74.093 g/mol = 2.667348 grams
∴ The theoretical yield of Ca(OH)₂ = 2.667348 grams
The percentage yield = (2.14 g)/(2.667348 grams) × 100 = 80.23%
Therefore, the option which is closest is option d. 81.1.
Answer:
Fe₂O₃ is the limiting reactant.
7.57 g of MgO are formed.
Explanation:
- 3 Mg + 1 Fe₂O₃ → 2 Fe + 3 MgO
First we <u>convert the given masses of both reactants into moles</u>, using their <em>respective molar masses</em>:
- 15.6 g Mg ÷ 24.305 g/mol = 0.642 mol Mg
- 10.0 g Fe₂O₃ ÷ 159.69 g/mol = 0.0626 mol Fe₂O₃
0.0626 moles of Fe₂O₃ would react completely with (3 * 0.0626 ) 0.188 moles of Mg. As there are more Mg moles than required, Mg is the reactant in excess; thus, <em>Fe₂O₃ is the limiting reactant</em>.
We now <u>calculate how many MgO moles are produced</u>, using the <em>number of moles of the limiting reactant</em>:
- 0.0626 mol Fe₂O₃ *
= 0.188 mol MgO
Finally we <u>convert moles of MgO into grams</u>:
- 0.188 mol MgO * 40.3 g/mol = 7.57 g