The actual yield is 43 g Cl₂.
The <em>limiting reactant was MnO₂</em> because it gave the smaller mass of Cl₂.
∴ The <em>theoretical yield</em> is 60.25 g Cl₂.
% yield = actual yield/theoretical yield × 100 %
Actual yield = theoretical yield × (% yield/100 %) = 60.25 g × (72 %/100%) = 43 g
You should put a picture or finish the question
Answer:
0.641 moles of ethane
Explanation:
Based on the equation:
C2H6(g) + 7/2O2(g) → 2CO2(g) + 3H2O(l)
We can determine ΔH of reaction using Hess's law. For this equation:
<em>Hess's law: ΔH products - ΔH reactants</em>
ΔH = {2ΔHCO2 + 3ΔHH2O} - {ΔHC2H6}
<em>Pure monoatomic substances have a ΔH = 0kJ/mol; ΔHO2 = 0kJ/mol</em>
<em />
ΔH = {2*-393.5kJ/mol + 3*-285.8kJ/mol} - {-84.7kJ/mol}
ΔH = -1559.7kJ/mol
That means when 1 mole of ethane is in combustion there are released 1559.7kJ of heat. To produce 1.00x10³kJ there are needed:
1.00x10³kJ * (1mole ethane / 1559.7kJ) =
<h3>0.641 moles of ethane</h3>
This is false, a compound does not have to contain both cations and anions if ions are present.
