Answer:
The answer to your question is: 70.8%
Explanation:
Data
Al₂O₃ = 60 g
C = 30 g
CO = gas
Al = 22.5 g
MW Al₂O₃ = 102 g
MW C = 12 g
MW Al = 54 g
Reaction
Al₂O₃ + 3C ⇒ 3 CO + 2 Al
Limiting reactant
102 g of Al₂O₃ -------------- 54 g Al
60 g -------------- x
x = 31.8 g
36 g of C ------------------ 54 g of Al
30 g of C ------------------ x
x = 45 g of Al
Limiting reactant = Al₂O₃
Percent yield = 
Percent yield = 70.75 %
Answer:
the answer is D the dominant over powers the resecive traits
The balanced chemical reaction is:
<span>C6H12 + 9O2 = 6CO2 + 6H2O
We are given the amount of </span><span>C6H12 to be used for the reaction. THis will be the starting point for the calculations.
1.900 g </span>C6H12 ( 1 mol C6H12/ 84.18 g C6H12) ( 6 mol CO2 / 1 mol <span>C6H12) (44.01 g /mol ) = 5.96 g CO2
</span>1.900 g C6H12 ( 1 mol C6H12/ 84.18 g C6H12) ( 6 mol H2O / 1 mol <span>C6H12) (18.02 g /mol ) = 2.44 g H2O
</span>
Answer: The enthalpy of combustion, per mole, of butane is -2657.4 kJ
Explanation:
The balanced chemical reaction is,
The expression for enthalpy change is,
Putting the values we get :
2 moles of butane releases heat = 5314.8 kJ
1 mole of butane release heat = 
Thus enthalpy of combustion per mole of butane is -2657.4 kJ