How many liters of chlorine gas at 25C and 0.950 atm can be produced by the reaction of 24.0 g of MnO2 with hydrochloric acid? The products include manganese (II) Chloride, water, and chlorine.
You have posted in the last day or so several problems that are similar. They are stoichiometry problems and all of those are worked the same way i a 4-step process. Here are the steps with this problem. I suggest you print and save this.
1. Write and balance the equation.
MnO2 + 4HCl ==> MnCl2 + Cl2 + 2H2O
2. Convert what you have (in this case 24.0 g MnO2) to mols. mol = grams/molar mass = ?
3. Using the coefficients in the balanced equation, convert mols of what you have (MnO2) to mols of what you want (in this case mols of what you want is mols Cl2). You do it this way.
mols Cl2 = mols MnO2 x (1 mol Cl2/1 mol MnO2) = mols MnO2 x 1 = ?; mols Cl2 = mols MnO2.
4. Now convert mols Cl2 to whatever unit you wish.
a. Most problems ask for grams Cl2. That is grams = mols x molar mass = ?
b. This problem asks for volume. So use PV = nRT, substitute P and T from the problem and solve for volume in liters. Don't forget T must be in kelvin.
Post your work if you get stuck.
Thanks. I see what I did wrong now, the T is 298 K not 273 K.
To determine the number of liters of chlorine gas produced, we need to follow these steps:
Step 1: Write the balanced equation for the reaction:
MnO2 + 4 HCl -> MnCl2 + 2 H2O + Cl2
Step 2: Determine the limiting reactant:
To find the limiting reactant, we need to compare the amount of MnO2 with the amount of HCl. We'll use the molar masses to convert grams to moles:
24.0 g MnO2 * (1 mol MnO2 / 86.94 g MnO2) = 0.276 mol MnO2
0.276 mol MnO2 * (4 mol HCl / 1 mol MnO2) = 1.10 mol HCl
Step 3: Convert limiting reactant to moles of chlorine gas:
From the balanced equation, we can see that 1 mole of MnO2 produces 1 mole of Cl2 gas. Therefore, 1.10 moles of HCl will also produce 1.10 moles of Cl2 gas.
Step 4: Use the ideal gas law to calculate the volume of Cl2 gas:
The ideal gas law relates the pressure (P), volume (V), temperature (T), and number of moles (n) of a gas:
PV = nRT
We need to rearrange the equation to solve for volume (V):
V = nRT / P
Given:
Temperature (T) = 25°C = 298 K
Pressure (P) = 0.950 atm
Number of moles (n) = 1.10 mol
Ideal gas constant (R) = 0.0821 L·atm/(K·mol)
V = (1.10 mol * 0.0821 L·atm/(K·mol) * 298 K) / 0.950 atm
V ≈ 30.77 liters (rounded to two decimal places)
Therefore, approximately 30.77 liters of chlorine gas can be produced.
To determine the number of liters of chlorine gas produced, we need to use the given information and balance the chemical equation. Here’s how you can do it step by step:
1. Write the balanced chemical equation:
MnO2 + 4 HCl → MnCl2 + 2 H2O + Cl2
2. Calculate the number of moles of MnO2:
Given mass of MnO2 = 24.0 g
Molar mass of MnO2 = 86.94 g/mol
Moles of MnO2 = Mass / Molar mass = 24.0 g / 86.94 g/mol
3. Use the stoichiometry of the balanced equation to find the moles of chlorine gas (Cl2):
From the balanced equation, we see that the mole ratio between MnO2 and Cl2 is 1:1.
Therefore, the moles of Cl2 produced will be equal to the moles of MnO2.
4. Use the Ideal Gas Law to calculate the volume of chlorine gas (Cl2):
PV = nRT
P = 0.950 atm (given)
V = Volume of Cl2 (what we need to find)
n = Moles of Cl2 (from step 3)
R = Ideal gas constant = 0.0821 L·atm/(mol·K)
T = 25°C = 25°C + 273.15 K = 298.15 K
Rearrange the equation to solve for V:
V = (nRT) / P
5. Plug in the values and solve for V:
V = (nRT) / P = (moles of Cl2) x (0.0821 L·atm/(mol·K)) x (298.15 K) / 0.950 atm
By following these steps with the given mass of MnO2 (24.0 g) and the balanced equation, you can calculate the volume of chlorine gas produced (in liters) when the reaction occurs at 25°C and a pressure of 0.950 atm.