Consider the following equation, which represents the combustion of hydrogen sulfide.
2H2S(g)+3O2(g)�¨2SO2(g)+2H2O(g)
What volume, in liters, of SO2 is formed when 120L of H2S(g) is burned? Assume that both gases are measured under the same conditions.
You should find the arrow key on your computer and use it.
When using gases at the same T and P one can use a shortcut procedure in which the volume (in L) is used directly as if it were mols.
So all you need to do is to convert 120 L H2S to SO2.
120 L SO2 x (2 mols SO2/2 mols H2S) \ 120 x 2/2 = 120 L SO2 formed.
To find the volume of SO2 formed when 120L of H2S(g) is burned, we need to use the stoichiometry of the balanced equation.
The equation tells us that 2 moles of H2S react to produce 2 moles of SO2. Therefore, there is a 1:1 mole ratio between H2S and SO2.
First, we need to convert 120L of H2S(g) into moles. We can use the ideal gas law to do this.
PV = nRT
Assuming constant temperature and pressure, we can rearrange the equation to solve for moles (n):
n = PV / RT
Next, we use the molar ratio to determine the moles of SO2 formed. Since the ratio of moles of H2S to SO2 is 1:1, the moles of SO2 formed will also be equal to the moles of H2S.
Finally, we can convert the moles of SO2 into volume using the ideal gas law:
V = nRT / P
By plugging in the values for moles of SO2, known temperature, pressure, and gas constant, we can find the volume of SO2 formed in liters.
Make sure to always double-check the units and ensure they are consistent throughout the calculation.