If that in the middle between the 10 and 24 is a caret, then I believe this is the work:
3.6 x 10^24 atoms Zn / 6.022x10^23 atom/mol Zn = 6 mol Zn
6 mol Zn x 65.36 g/mol Zn = 390 g Zn if you carry two significant figures with rounding at each step.
I believe that is correct as all the units cancel properly.
The question is incomplete. The complete question is :
C. Balance these fossil-fuel combustion reactions. (1 point)
C8H18(g) + 12.5O2(g) → ____CO2(g) + 9H2O(g) + heat
CH4(g) + ____O2(g) → ____CO2(g) + ____H2O(g) + heat
C3H8(g) + ____O2(g) → ____CO2(g) + ____H2O(g) + heat
C6H6(g) + ____O2(g) → ____CO2(g) + ____H2O(g) + heat
Solution :
C8H18(g) + 12.5O2(g) → __8__CO2(g) + 9H2O(g) + heat
When 1 part of octane reacts with 12.5 parts of oxygen, it gives 8 parts of carbon dioxide and 9 parts of water along with liberation of energy.
CH4(g) + __2__O2(g) → __1__CO2(g) + __2__H2O(g) + heat
When 1 part of methane reacts with 2 parts of oxygen, it gives 1 part of carbon dioxide and 2 parts of water along with liberation of energy.
C3H8(g) + __5__O2(g) → __3__CO2(g) + __4__H2O(g) + heat
When 1 part of propane reacts with 5 parts of oxygen, it gives 3 part of carbon dioxide and 4 parts of water along with liberation of energy.
C6H6(g) + __1/2__O2(g) → __6__CO2(g) + __3__H2O(g) + heat
When 1 part of propane reacts with 1/2 parts of oxygen, it gives 6 part of carbon dioxide and 3 parts of water along with liberation of energy.
Density is defined as mass per unit volume
therefore density = mass ÷ volume
volume = mass÷density
volume in ml = 17 ÷ 3.291
Answer:
Oxygen is the limiting reactant.
Explanation:
Based on the reaction:
C₁₂H₂₂O₁₁ + 12O₂ → 12CO₂ + 11H₂O
<em>1 mole of sucrose reacts with 12 moles of oxygen to produce 12 moles of CO₂ and 11 moles of H₂O.</em>
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10.0g of sucrose (Molar mass: 342.3g /mol) are:
10.0g C₁₂H₂₂O₁₁ × (1mole / 342.3g) = 0.0292 moles of C₁₂H₂₂O₁₁
And moles of 10.0g of oxygen (Molar mass: 32g/mol) are:
10.0g O₂ × (1mole / 32g) = 0.3125 moles of O₂
For a complete reaction of 0.0292 moles of C₁₂H₂₂O₁₁ you need (knowing 12 moles of oxygen react per mole of sucrose):
0.0292 moles of C₁₂H₂₂O₁₁ × (12 moles O₂ / 1 mole C₁₂H₂₂O₁₁) = <em>0.3504 moles of O₂</em>
As you have just 0.3125 moles of O₂, <em>oxygen is the limiting reactant.</em>
