a temperature and pressure of 23.5°C and 1.05 atm, determine the volume of sulfur dioxide produced when 150.0 g of dihydrogen sulfide react in excess oxygen.
To determine the volume of sulfur dioxide (SO2) produced when 150.0 g of dihydrogen sulfide (H2S) reacts in excess oxygen, we need to use the stoichiometry of the balanced chemical equation.
First, let's write the balanced chemical equation for the reaction:
2 H2S + 3 O2 -> 2 SO2 + 2 H2O
From the balanced equation, we can see that 2 moles of H2S react to produce 2 moles of SO2. Therefore, we need to convert the mass of H2S to moles.
To do this, we need to know the molar mass of H2S. The molar mass of hydrogen (H) is approximately 1.01 g/mol, and the molar mass of sulfur (S) is approximately 32.07 g/mol. Adding them up, the molar mass of H2S is:
(2 * molar mass of H) + molar mass of S
= (2 * 1.01 g/mol) + 32.07 g/mol
= 2.02 g/mol + 32.07 g/mol
= 34.09 g/mol
Now, we can use the given mass of H2S (150.0 g) and the molar mass of H2S to calculate the number of moles of H2S:
Number of moles = mass / molar mass
= 150.0 g / 34.09 g/mol
= 4.399 mol (rounded to 3 decimal places)
Since 2 moles of H2S produce 2 moles of SO2, we can conclude that 4.399 moles of H2S will produce 4.399 moles of SO2.
Now, let's use the ideal gas law to find the volume of SO2 produced at the given temperature and pressure. The ideal gas law equation is:
PV = nRT
Where:
P = Pressure (1.05 atm)
V = Volume (Unknown)
n = Number of moles (4.399 moles)
R = Ideal gas constant (0.0821 L.atm/mol.K)
T = Temperature in Kelvin (23.5 °C + 273.15 = 296.65 K)
Rearranging the equation to solve for V:
V = (nRT) / P
Substituting the given values:
V = (4.399 mol * 0.0821 L.atm/mol.K * 296.65 K) / 1.05 atm
Calculating:
V = 0.996 L (rounded to 3 decimal places)
Therefore, the volume of sulfur dioxide produced when 150.0 g of dihydrogen sulfide reacts in excess oxygen at a temperature of 23.5 °C and pressure of 1.05 atm is approximately 0.996 L.