Question

Pentane gas reacts with oxygen gas to give carbon dioxide gas and water vapor (gas). If you mix pentane and oxygen in the correct stoichiometric ratio, and if the total pressure of the mixture is 180 mm Hg, what are the partial pressures of pentane ( mmHg) and oxygen ( mm Hg)? If the temperature and volume do not change, what is the pressure of the water vapor ( mm Hg) after reaction?

Answer 1

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;

$C_5H_{12}_{(g)}+8O_{2(g)}$  ⇒  $5CO_{2(g)}+6H_2O_{(g)}$

Given that;

The total pressure $(P_{Total})$= 180 mmHg

From the equation above; we can see that;

the numbers of moles of  $C_5H_{12}_{(g)$  i.e $(M_C__5}_H__1_2}__(_g_)})$ = 1

the numbers of moles of $O_2$ i.e $(M_O__2})$

∴ Using the formula  for Dalton Law of Partial Pressure;

The partial pressure for pentane can be calculated as:

$P_C__5}_H__{12}$ =  $(\frac{M_C_5_H_1_2}{M_C_5_H_1_2+M_O__2}}) *P_{Total}$

= $\frac{1}{8+1}(180)$

= 0.1111 × (180)mmHg

= 19.98 mmHg

≅ 20 mm Hg

The partial pressure for oxygen can be calculated as:

$P_O___2}$ = $(\frac{M_O__2}{M_C_5_H_1_2+M_O__2}}) *P_{Total}$

= $\frac{8}{1+8} (180)$

= 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.

                                  $C_5H_{12}_{(g)}+8O_{2(g)}$  ⇒  $5CO_{2(g)}+6H_2O_{(g)}$

Initial (mmHg)               20            160                 0             0

change                          -20           -8 (20)        5 (20)         6(20)

Equilibrium (mmHg)      0                0               100             120

$P_{Total}$  $= P_C_O__2}+P_H__2O$

= (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;

$P_H___2}O}$  $= (\frac{M_H__2_O}{M_H__2_O+M_C_O__2}}}) *P_{Total}$

$= (\frac{5}{5+6}}}) *220$

= 0.5454 (220) mmHg

= 119.98 mmHg

≅ 120 mmHg

∴  the pressure of the water vapor(H₂O mmHg) after reaction = 120 mmHg