If a 0.500 mole of carbon disulfide reacts with oxygen completely, what would be the total volume of the products at STP?
CS2(l) + 3 O2(g) → CO2(g) + 2 SO2(g)
each mole of CS2 produces 3 moles of gas
... each moles of gas has a volume of 22.4 L
To find the total volume of the products at STP, we can use the ideal gas law, which states:
PV = nRT
Where:
P = Pressure
V = Volume
n = Number of moles
R = Gas constant
T = Temperature
Since the question specifies that we are at STP, we can assume that the temperature (T) is 273.15 K and the pressure (P) is 1 atm. The gas constant (R) is typically given as 0.0821 L·atm/(mol·K) in this context.
First, we need to determine the number of moles of each product. From the balanced chemical equation, we can see that 1 mole of carbon disulfide (CS2) reacts to form 1 mole of carbon dioxide (CO2) and 2 moles of sulfur dioxide (SO2). Therefore, 0.500 moles of CS2 would produce 0.500 moles of CO2 and 1.000 mole of SO2.
Next, we can use the ideal gas law to find the volume of each gas produced. Let's start with carbon dioxide:
n = 0.500 moles
P = 1 atm
T = 273.15 K
R = 0.0821 L·atm/(mol·K)
Rearranging the ideal gas law equation to solve for V:
V = nRT / P
V(CO2) = (0.500 mol)(0.0821 L·atm/(mol·K))(273.15 K) / 1 atm
V(CO2) = 11.2 L
So, 0.500 mole of carbon disulfide would produce 11.2 liters of carbon dioxide at STP.
Now let's calculate the volume of sulfur dioxide:
n = 1.000 mole
P = 1 atm
T = 273.15 K
R = 0.0821 L·atm/(mol·K)
V(SO2) = (1.000 mol)(0.0821 L·atm/(mol·K))(273.15 K) / 1 atm
V(SO2) = 22.4 L
Therefore, 0.500 mole of carbon disulfide would produce 22.4 liters of sulfur dioxide at STP.
To find the total volume of the products, we can simply add the volume of carbon dioxide and sulfur dioxide together:
Total Volume = V(CO2) + V(SO2)
Total Volume = 11.2 L + 22.4 L
Total Volume = 33.6 L
So, the total volume of the products at STP would be 33.6 liters.