A closed gas cylinder contains 0.500 mole H2, 0.300 mole O2, 1.200 mole CO2 at 25 °C and pressures 2.00 atm. Calculate the volume of the cylinder.
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Answer to Question #123197 in General Chemistry for Dyana
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Question #123197
A closed gas cylinder contains 0.500 mole H2, 0.300 mole O2, 1.200 mole CO2 at 25 °C and pressures 2.00 atm. Calculate the volume of the cylinder. Then, calculate the partial pressure of O2 inside the cylinder in question.
((R=0.082 L atm mol^-1 K^-1)
Expert's answer
Solution.
n_{H_2}=0.500mol;n
H
2
=0.500mol;
n_{O_2}=0.300mol;n
O
2
=0.300mol;
n_{CO_2}=1.200mol;n
CO
2
=1.200mol;
T=25^oC=298K;T=25
o
C=298K;
P_{total}=2 atm;P
total
=2atm;
P_{total}V=n_{total}RT;P
total
V=n
total
RT;
V=\dfrac{n_{total}RT}{P_{total}};V=
P
total
n
total
RT
;
n_{total}=n_{H_2}+n_{O_2}+n_{CO_2};n
total
=n
H
2
+n
O
2
+n
CO
2
;
V=\dfrac{(0.500+0.300+1.200)mol\sdot0.082L \sdot atm \sdot mol^{-1}K^{-1}\sdot298K}{2atm}=24.44L;V=
2atm
(0.500+0.300+1.200)mol⋅0.082L⋅atm⋅mol
−1
K
−1
⋅298K
=24.44L;
P_{O_2}=n_{O_2}\dfrac{RT}{V};P
O
2
=n
O
2
V
RT
;
P_{O_2}=0.300mol \dfrac{0.082L\sdot atm\sdot mol^{-1}\sdot K^{-1}\sdot298K}{24.44L}=0.3atm;P
O
2
=0.300mol
24.44L
0.082L⋅atm⋅mol
−1
⋅K
−1
⋅298K
=0.3atm;
Answer: V=24.44L;V=24.44L;
P_{O_2}=0.3atm;P
O
2
=0.3atm;