In the reaction: MnO2 + 4 HCl ---> MnCl2 + Cl2 + 2 H2O
Calculate:
The volume of Cl2 produced from the reaction of 10 grams of HCl with excess MnO2?
Here is a worked example of a stoichiometric problem. Just follow the steps. After you find moles Cl2, moles x 22.4 L/mol will give you the volume.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To calculate the volume of Cl2 produced from the reaction of 10 grams of HCl with excess MnO2, we need to use the stoichiometry of the reaction.
1. Convert the mass of HCl to moles:
The molar mass of HCl (hydrochloric acid) is approximately 36.46 g/mol.
10 grams of HCl is equal to 10 g / 36.46 g/mol = 0.274 moles of HCl.
2. Use the stoichiometric coefficients from the balanced equation to find the moles of Cl2 produced:
From the balanced equation: 1 mole of HCl reacts to produce 1 mole of Cl2.
Therefore, the 0.274 moles of HCl will produce 0.274 moles of Cl2.
3. Convert moles of Cl2 to volume at standard temperature and pressure (STP):
At STP (0 degrees Celsius and 1 atmosphere), 1 mole of any ideal gas occupies 22.4 liters.
Thus, 0.274 moles of Cl2 will occupy 0.274 moles * 22.4 L/mol = 6.13 liters of Cl2.
Therefore, the volume of Cl2 produced from the reaction of 10 grams of HCl with excess MnO2 is approximately 6.13 liters.
To calculate the volume of Cl2 produced from the given reaction, we first need to determine the number of moles of HCl using its molar mass.
1. Determine the molar mass of HCl:
- H: 1 g/mol
- Cl: 35.5 g/mol
Molar mass of HCl = 1 g/mol + 35.5 g/mol = 36.5 g/mol
2. Calculate the moles of HCl:
Moles = Mass / Molar mass
Moles of HCl = 10 g / 36.5 g/mol = 0.274 moles
3. According to the balanced chemical equation, the stoichiometric ratio between HCl and Cl2 is 4:1. This means for every 4 moles of HCl, we get 1 mole of Cl2.
4. Determine the moles of Cl2:
Moles of Cl2 = Moles of HCl / Stoichiometric ratio
Moles of Cl2 = 0.274 moles / 4 = 0.0685 moles
5. Finally, we can use the ideal gas law to calculate the volume of Cl2:
PV = nRT
V = (nRT / P)
Where:
n = moles of Cl2
R = ideal gas constant (0.0821 L·atm/(K·mol))
T = temperature (in Kelvin)
P = pressure (in atm)
Since the temperature and pressure are not given, let's assume standard temperature and pressure (STP):
T = 273 K
P = 1 atm
V = (0.0685 mol * 0.0821 L·atm/(K·mol) * 273 K) / 1 atm
V ≈ 1.68 L
So, the volume of Cl2 produced from the reaction of 10 grams of HCl with excess MnO2 is approximately 1.68 liters.