Hydrogen sulfide gas is combusted with oxygen gas to produce sulfur dioxide gas and water vapour. If there is 19.3 L of oxygen gas, what maximum volume of hydrogen sulfide gas could be combusted?
its hard to find
To determine the maximum volume of hydrogen sulfide gas that could be combusted, we need to use the balanced chemical equation for the combustion reaction between hydrogen sulfide and oxygen.
The balanced chemical equation is:
2 H2S(g) + 3 O2(g) -> 2 SO2(g) + 2 H2O(g)
From the balanced chemical equation, we can see that for every 2 moles of hydrogen sulfide gas (H2S), we need 3 moles of oxygen gas (O2) to completely combust the hydrogen sulfide.
First, we need to convert the volume of oxygen gas (19.3 L) into moles using the ideal gas law equation:
PV = nRT
Where:
P is the pressure (which we can assume to be constant)
V is the volume (19.3 L in this case)
n is the number of moles (what we want to find)
R is the ideal gas constant (0.0821 L·atm/(mol·K))
T is the temperature (assumed to be constant)
Rearranging the equation to solve for n:
n = PV/RT
Substituting the values:
n = (19.3 L)(1 atm) / (0.0821 L·atm/(mol·K) × T)
For the volume of hydrogen sulfide gas (H2S), we can use the mole ratio from the balanced chemical equation to determine the moles of H2S needed to react with the moles of O2:
2 moles of H2S / 3 moles of O2 = n moles of H2S / (n moles of O2)
Simplifying,
2 / 3 = n moles of H2S / [(19.3 L)(1 atm) / (0.0821 L·atm/(mol·K) × T)]
Solving for n moles of H2S:
n moles of H2S = (2/3) × [(19.3 L)(1 atm) / (0.0821 L·atm/(mol·K) × T)]
Finally, to determine the volume of H2S gas, we multiply the moles of H2S by its molar volume at STP (Standard Temperature and Pressure), which is 22.4 L/mol.
Volume of H2S gas = (n moles of H2S) × (22.4 L/mol)
Note: Make sure to substitute the correct temperature value for T in the calculation.