How many liters of Chlorine at STP will react with 35L of 02 at STP given the following reaction?
2Cl2(g) + 7O2(g) - 2Cl2O7(l)
10 L Cl2
You can use regular stoichiometry to solve this problem or you can take a shortcut. The shortcut works this way. Instead of converting to moles, if all of the reactants and products are in the gaseous form, volume may be used as if volume = moles.
Therefore, use the coefficients to convert L of one reagent to L of another.
35L O2 x (2 moles Cl2/7 moles O2) = ??L Cl2.
10
So, 10 L of Cl2 at STP will react with 35 L of O2 at STP according to the given reaction.
Well, it looks like you're trying to react Chlorine (Cl2) with Oxygen (O2) to produce Chlorine Heptoxide (Cl2O7). According to the balanced equation, it takes 2 moles of Cl2 to react with 7 moles of O2.
Now, given that we have 35 liters of O2 at STP, we can use the molar volume of gases at STP (22.4 L/mol) to determine the number of moles of O2 present:
35 L O2 / 22.4 L/mol = approximately 1.56 moles of O2
Since the ratio of Cl2 to O2 is 2:7, we can set up a proportion to find out how many moles of Cl2 will be required:
2 moles Cl2 / 7 moles O2 = x moles Cl2 / 1.56 moles O2
Simplifying the proportion, we find:
x ≈ (2/7) × 1.56 ≈ 0.45 moles of Cl2
Now, we can use the molar volume to convert moles to liters at STP:
0.45 moles Cl2 × 22.4 L/mol = approximately 10.08 liters of Cl2
So, it looks like you'll need approximately 10.08 liters of Chlorine at STP to react with 35 liters of Oxygen at STP. But hey, chemistry isn't always so straightforward. It's like a wild circus sometimes!
To determine the number of liters of chlorine gas (Cl2) at STP that will react with 35 liters of oxygen gas (O2) at STP, we need to use the stoichiometry of the balanced chemical equation.
The balanced chemical equation is:
2Cl2(g) + 7O2(g) → 2Cl2O7(l)
From the balanced equation, we can see that 2 moles of Cl2 react with 7 moles of O2 to produce 2 moles of Cl2O7. This mole ratio can be used to calculate the number of moles of Cl2 we need for the given amount of O2.
First, let's convert the volume of O2 from liters to moles using the ideal gas law:
PV = nRT
At STP (Standard Temperature and Pressure), the values are:
P = 1 atm
V = 35 L
T = 273.15 K
R = 0.0821 L·atm/mol·K
Using the ideal gas law:
n = PV / RT
n = (1 atm) * (35 L) / (0.0821 L·atm/mol·K * 273.15 K)
n = 1.13 moles of O2
Now that we have the number of moles of O2, we can use the mole ratio from the balanced equation to find the number of moles of Cl2 needed.
According to the balanced equation, 2 moles of Cl2 react with 7 moles of O2. Therefore, we can set up a ratio:
2 moles Cl2 / 7 moles O2 = x moles Cl2 / 1.13 moles O2
Solving for x:
x = (2 moles Cl2 / 7 moles O2) * 1.13 moles O2
x = 0.32
Therefore, we need 0.32 moles of Cl2 to react with 35 liters of O2 at STP.
Finally, we can convert this mole quantity to liters of Cl2 using the ideal gas law again:
V = nRT / P
V = (0.32 moles) * (0.0821 L·atm/mol·K * 273.15 K) / (1 atm)
V ≈ 7.09 L
Hence, approximately 7.09 liters of Cl2 gas at STP will react with 35 liters of O2 gas at STP.