Answer:
1.12×10¹¹ kg of CO₂ are produced with 4.6×10¹⁰ L of isooctane
Explanation:
Let's state the combustion reaction:
C₈H₁₈ + 25/2O₂ → 8CO₂ + 9H₂O
Let's calculate the mass of isooctane that reacts.
Density = Mass / Volume
Density . Volume = Mass
First of all, let's convert the volume in L to mL, so we can use density.
4.6×10¹⁰ L . 1000 mL / 1L = 4.6×10¹³ mL
0.792 g/mL . 4.6×10¹³ mL = 3.64 ×10¹³ g
This mass of isooctane reacts to produce CO₂ and water, so let's determine the moles of reaction
3.64 ×10¹³ g . 1mol / 114 g = 3.19×10¹¹ mol
Ratio is 1:8 so 1 mol of isooctane can produce 8 moles of dioxide
Therefore 3.19×10¹¹ mol would produce (3.19×10¹¹ mol . 8) = 2.55×10¹² moles of CO₂
Now, we can determine the mass of produced CO₂ by multipling:
moles . molar mass
2.55×10¹² mol . 44 g/mol = 1.12×10¹⁴ g of CO₂
If we convert to kg 1.12×10¹⁴ g / 1000 = 1.12×10¹¹ kg
increases in solution resulting removal of
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. Hence, according to Le-chatelier principle, equilibrium will shift towards right. So more
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increases in solution resulting shifting of equilibrium towards left in accordance with Le-chatelier principle. So less