8.8 g of iron(II) sulphide reacts completely with excess dilute hydrochloric acid. Calculate the volume of hydrogen sulphide evolved, measured at 18 º C and 729 mmHg pressure.
1. Calculate moles of H2S from Rxn
2. Cal. Vol H2S at STP = moles H2S x 22.4 l/mole at STP
3. Convert Vol H2S at STP to Vol at 18C & 729 mmHg using Ideal Gas Law
Ans => 2.489 Liters H2S
To calculate the volume of hydrogen sulphide evolved, we need to use the Ideal Gas Law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = moles of gas
R = ideal gas constant (0.0821 L·atm / K·mol)
T = temperature (in Kelvin)
First, let's convert the given pressure and temperature to the required units:
Pressure:
Given: 729 mmHg
To convert to atm, divide by 760 mmHg/atm:
729 mmHg ÷ 760 mmHg/atm ≈ 0.96 atm
Temperature:
Given: 18 ºC
To convert to Kelvin, add 273.15:
18 ºC + 273.15 ≈ 291.15 K
Now, we need to determine the number of moles of hydrogen sulphide (H2S) produced. We can do this by using the balanced chemical equation for the reaction between iron(II) sulphide (FeS) and hydrochloric acid (HCl):
FeS + 2HCl → FeCl2 + H2S
From the equation, we see that for each mole of FeS, we get one mole of H2S. The molar mass of FeS is 87.9 g/mol. Therefore, we can calculate the number of moles of FeS:
Moles of FeS = mass of FeS / molar mass of FeS
Moles of FeS = 8.8 g / 87.9 g/mol ≈ 0.100 mol
Since the reaction is complete and FeS is in excess, the number of moles of H2S formed will also be approximately 0.100 mol.
Now we have all the information needed to calculate the volume of H2S:
PV = nRT
V = (nRT) / P
V = (0.100 mol)(0.0821 L·atm / K·mol)(291.15 K) / 0.96 atm
V ≈ 2.43 L
Therefore, the volume of hydrogen sulphide evolved, measured at 18 ºC and 729 mmHg pressure, is approximately 2.43 liters.