1) Balanced equation
C3H8 + 5O2 -> 3 CO2 + 4 H2O
2) 0.700 L C3H8
Given the pressure and temperature do not change, the molar ratio is equivalent to volume ratio
1molC3H8 / 5 mol O2 => 1 L C3H8 / 5 L O2
0.700 L C3H8 / x L O2 = 1 L C3H8 / 5 L O2 => x = 0.700 L C3H8 * 5 L O2 / 1 L C3H8
x = 3.500 L O2
3) CO2 produced
1 L C3H8 / 3 L CO2 = 0.700 L C3H8 / x L CO2 =>
x = 0.700 L C3H8 * 3 L CO2 / 1 L C3H8 = 2.100 L CO2
4) Water vapor produced
1) 1 L C3H8 / 4 L H2O = 0.700 LC3H8 / x L H2O =>
x = 0.700 L C3H8 * 4 L H20 / 1 L C3H8 = 2.800 L H2O
Just multiple 19.3 by 15 because cm^3 will cancel outand you are left with g with represents mass
Answer:
Heated water begins circulating in a fish tank.
Explanation:
Answer:
The equilibrium concentration of hydrogen gas is 0.0010 M.
Explanation:
The equilibrium constant of the reaction = 
}
Moles of hydrogen sulfide = 0.31 mol
Volume of the container = 4.1 L
![[concentration]=\frac{moles}{volume (L)}](https://tex.z-dn.net/?f=%5Bconcentration%5D%3D%5Cfrac%7Bmoles%7D%7Bvolume%20%28L%29%7D)
![[H_2S]=\frac{0.31 mol}{4.1 L}=0.076 M](https://tex.z-dn.net/?f=%5BH_2S%5D%3D%5Cfrac%7B0.31%20mol%7D%7B4.1%20L%7D%3D0.076%20M)
Initially
0.076 M
At equilibrium
(0.076-2x) 2x x
The expression of an equilibrium constant :
![K_c=\frac{[H_2]^2[S_2]}{[H_2S]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BH_2%5D%5E2%5BS_2%5D%7D%7B%5BH_2S%5D%5E2%7D)
Solving for x:
x = 0.00051
The equilibrium concentration of hydrogen gas:
![[H_2]=2x=2\times 0.00051 M=0.0010 M](https://tex.z-dn.net/?f=%5BH_2%5D%3D2x%3D2%5Ctimes%200.00051%20M%3D0.0010%20M)
