Answer:
The partial pressure for pentane = 20 mmHg
The partial pressure for oxygen = 160 mmHg
The pressure of the water vapor(H₂O mmHg) after reaction = 120 mmHg
Explanation:
The balanced equation for the reaction is written below as;
⇒ 
Given that;
The total pressure
= 180 mmHg
From the equation above; we can see that;
the numbers of moles of
i.e
= 1
the numbers of moles of
i.e 
∴ Using the formula for Dalton Law of Partial Pressure;
The partial pressure for pentane can be calculated as:
= 
= 
= 0.1111 × (180)mmHg
= 19.98 mmHg
≅ 20 mm Hg
The partial pressure for oxygen can be calculated as:
= 
= 
= 0.889 * (180) mmHg
= 160 mmHg
If the temperature and volume do not change, what is the pressure of the water vapor ( mm Hg) after reaction?
NOW, From the equation above, After the reaction is complete;
5 moles of CO₂ and 6 moles of H₂O were produced.
⇒ 
Initial (mmHg) 20 160 0 0
change -20 -8 (20) 5 (20) 6(20)
Equilibrium (mmHg) 0 0 100 120

= (100 + 120) mmHg
=220 mmHg
numbers of moles of CO₂ = 5 moles
numbers of moles of H₂O = 6 moles
∴ To calculate the pressure of water vapor ( H₂O ) using Dalton's Law; we have;

= 0.5454 (220) mmHg
= 119.98 mmHg
≅ 120 mmHg
∴ the pressure of the water vapor(H₂O mmHg) after reaction = 120 mmHg