Answer:
Percent yield: 78.2%
Explanation:
Based on the reaction:
4Al + 3O₂ → 2Al₂O₃
<em>4 moles of Al produce 2 moles of Al₂O₃</em>
<em />
To find percent yield we need to find theoretical yield (Assuming a yield of 100%) and using:
(Actual yield (6.8g) / Theoretical yield) × 100
Moles of 4.6g of Al (Molar mass: 26.98g/mol) are:
4.6g Al × (1mol / 26.98g) = 0.1705 moles of Al.
As 4 moles of Al produce 2 moles of Al₂O₃, theoretical moles of Al₂O₃ obtained from 0.1705 moles of Al are:
0.17505 moles Al × (2 moles Al₂O₃ / 4 moles Al) = <em>0.0852 moles of Al₂O₃</em>,
In grams (Molar mass Al₂O₃ = 101.96g/mol):
0.0852 moles of Al₂O₃ × (101.96g / mol) =
<h3>8.7g of Al₂O₃ can be produced (Theoretical yield)</h3>
Thus, Percent yield is:
(6.8g / 8.7g) × 100 =
<h3>
78.2% </h3>
Balanced:
1. <span>Na2O + H2O ---> 2NaOH
2. </span><span>K2O + H2O ---> 2KOH
3. </span><span>MgO + H2O ---> Mg(OH)2
4. </span><span>CaO + H2O ---> Ca(OH)2
5. </span><span>SO2 + H2O ⇄ H2SO3
6. </span>SO3 + H2O ---> H2SO4
All except by 2 were balanced.
A. conduction
that is the correct answer
Carbon (24.00)(2)+hydrogen(1.01)(6)
percentage error=difference/actual value x 100
=0.534/13.0 x 100
=4.11%
