what volume of carbon dioxide is formed when 11 of CS2 reacts with 18 L O2 to produce carbon dioxide gas and sulfur dioxide gas at STP?
This is a limiting reagent problem. You know that because you have amounts given for BOTH reactants.
CS2 + 3O2 ==> CO2 + 2SO2
Is that 11 L CS2? If so you can take a shortcut and not convert L to mols but use L directly. As follows:
L CO2 formed will be
L CS2 x (1 mol CO2/1 mol CS2) = 11 x (1/1) = 11 L CO2 (If we had all of the O2 we needed.
If we had 18 L O2 the volume of CO2 formed will be
L O2 x (1 mol CO2/3 mols O2) = 18 x (1/3) = 6 L CO2 (if we had all of the CS2 need).
You get two different answers this way which means one of them is wrong; the correct answer in limiting reagent problems is ALWAYS the smaller value; therefore, 6L CO2 will be the volume of CO2 formed at STP.
To find the volume of carbon dioxide formed when 11 L of CS2 reacts with 18 L of O2 at STP, we need to balance the chemical equation and use the stoichiometry.
Here is the balanced equation for the reaction:
CS2 + 3O2 -> CO2 + 2SO2
From the balanced equation, we can see that for every 1 mole of CS2, it reacts with 3 moles of O2 to produce 1 mole of CO2.
Step 1: Calculate the number of moles of CS2:
11 L of CS2 is given. However, since the volume is not at STP, we need to convert it to moles using the ideal gas law equation:
PV = nRT
Assuming the CS2 is at room temperature (around 298 K) and 1 atm of pressure, the value of R is 0.0821 L·atm/(mol·K).
n(CS2) = (PV)/(RT)
= (1 atm × 11 L) / (0.0821 L·atm/(mol·K) × 298 K)
= 5.0 moles
Step 2: Using the stoichiometry of the balanced equation, we can determine the moles of CO2 produced. Since every mole of CS2 produces 1 mole of CO2, we have:
n(CO2) = 5.0 moles
Step 3: Convert the moles of CO2 to volume at STP.
1 mole of any ideal gas occupies 22.4 L at STP. Therefore, 5.0 moles of CO2 will occupy:
V(CO2) = n(CO2) × 22.4 L/mol
= 5.0 moles × 22.4 L/mol
= 112 L
So, when 11 L of CS2 reacts with 18 L of O2, the volume of carbon dioxide formed at STP is 112 L.
To find the volume of carbon dioxide formed, we need to use the balanced chemical equation for the reaction between CS2 and O2.
The balanced equation is as follows:
CS2 + 3O2 -> CO2 + 2SO2
From the equation, we can see that 1 mole of CS2 reacts with 3 moles of O2 to produce 1 mole of CO2.
First, we need to convert the given amount of CS2 from grams to moles. We need the molar mass of CS2, which is approximately 76.14 g/mol.
11 g of CS2 * (1 mol / 76.14 g) = 0.144 mol of CS2
Now, let's use the mole ratio between CS2 and CO2 to find the moles of CO2 produced.
0.144 mol of CS2 * (1 mol CO2 / 1 mol CS2) = 0.144 mol of CO2
Since we are given the reaction is taking place at STP, we know that 1 mole of any gas occupies 22.4 liters at STP.
Therefore, 0.144 mol of CO2 will occupy:
0.144 mol CO2 * 22.4 L/mol = 3.22 L of CO2
So, when 11 g of CS2 reacts with 18 L of O2 at STP, the volume of carbon dioxide formed is approximately 3.22 liters.