What volume of hydrogen sulfide is required to produce 161 liters of sulfur dioxide according to the following reaction? (All gases are at the same temperature and pressure.)
hydrogen sulfide (g) + oxygen (g)----> water (l) + sulfur dioxide (g)
2H2S + 3O2 ==> 2H2O + 2SO2
161L SO2 x (2 mols H2S/2 mols SO2) = 161 x 2/2 = ?
To find the volume of hydrogen sulfide required to produce 161 liters of sulfur dioxide, we first need to determine the molar ratio between the two gases in the reaction.
The balanced chemical equation for the reaction is:
2H2S(g) + 3O2(g) → 2H2O(l) + 2SO2(g)
From the balanced equation, we can see that for every 2 moles of hydrogen sulfide (H2S), we produce 2 moles of sulfur dioxide (SO2).
Now, we can use the ideal gas law to calculate the volume of hydrogen sulfide.
The ideal gas law equation is:
PV = nRT
Where:
P = pressure (constant in this case)
V = volume
n = moles
R = ideal gas constant
T = temperature (constant in this case)
In this case, we want to find the volume of hydrogen sulfide (V), given the number of moles (n) of hydrogen sulfide.
To find the number of moles of hydrogen sulfide, we can use the equation:
n = (V * P) / (R * T)
Since the pressure, temperature, and gas constant are the same for both gases, we can say that:
n(H2S) / n(SO2) = V(H2S) / V(SO2)
Now we can set up the ratio:
n(H2S) / 2 = 161 L / 2
Simplifying the equation:
n(H2S) = 161 L
Therefore, the volume of hydrogen sulfide required to produce 161 liters of sulfur dioxide would also be 161 liters.