How many liters of chlorine as can be produced when 1.96 L of HCl react with excess O2 at STP
Show Work
A.0.49L
B.0.98L
C.1.96L
D.3.92L
You work this exactly like the problem with glucose and oxygen.
1. Write and balance the equation.
2. Use the coefficients to convert mols of what you have to mols of what you want.
3. Convert mols of what you want to either grams or volume.
g = mols x molar mass
volume = mols x 22.4 L/mol if at STP or
PV = nRT if not at STP.
To determine the amount of chlorine gas (Cl2) that can be produced when 1.96 L of hydrogen chloride (HCl) reacts with excess oxygen (O2) at STP (standard temperature and pressure), we need to use the balanced chemical equation for the reaction.
The balanced equation for the reaction between hydrogen chloride and oxygen is:
4 HCl + O2 -> 2 Cl2 + 2 H2O
From the balanced equation, we can see that 4 moles of HCl react to produce 2 moles of Cl2.
To find the number of moles of HCl, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
At STP, the pressure (P) is 1 atm, the volume (V) is 1.96 L, the ideal gas constant (R) is 0.0821 L.atm/mol.K, and the temperature (T) is 273 K.
Plugging in the values:
1 atm * 1.96 L = n * 0.0821 L.atm/mol.K * 273 K
1.96 = n * 22.414
n = 1.96 / 22.414
n ≈ 0.0875 moles of HCl
Since 4 moles of HCl react to produce 2 moles of Cl2, we can use a mole ratio to find the number of moles of Cl2 produced:
0.0875 moles HCl * (2 moles Cl2 / 4 moles HCl) = 0.04375 moles Cl2
Finally, we can use the ideal gas law equation to find the volume of Cl2 gas at STP:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
Plugging in the values:
1 atm * V = 0.04375 moles * 0.0821 L.atm/mol.K * 273 K
V = (0.04375 * 0.0821 * 273) / 1
V ≈ 0.9886 L
Therefore, the volume of chlorine gas produced is approximately 0.9886 L.
So the correct answer choice is B. 0.98 L.