How many liters of chlorine gas can be produced when 0.98 L of HCl react with excess O2 at STP?
To find out how many liters of chlorine gas can be produced when 0.98 L of HCl reacts with excess O2 at STP (Standard Temperature and Pressure), we need to use the balanced chemical equation for the reaction between HCl and O2.
The balanced chemical equation is:
4 HCl + O2 -> 2 Cl2 + 2 H2O
From the equation, we can see that 4 moles of HCl react to produce 2 moles of Cl2.
To calculate the number of moles of HCl in 0.98 L of HCl, we'll need to use the ideal gas law:
PV = nRT
Where:
P is the pressure (which is equal to 1 atm at STP)
V is the volume of the gas (0.98 L)
n is the number of moles
R is the ideal gas constant (0.0821 L·atm/(mol·K))
T is the temperature (which is equal to 273.15 K at STP)
Rearranging the ideal gas law equation to solve for n, we get:
n = PV / RT
n = (1 atm) x (0.98 L) / (0.0821 L·atm/(mol·K) x 273.15 K)
This gives us the number of moles of HCl in 0.98 L.
Next, we use the stoichiometry of the balanced chemical equation to determine the number of moles of Cl2 produced.
From the balanced chemical equation, we know that 4 moles of HCl react to produce 2 moles of Cl2.
Therefore, if we have "n" moles of HCl, we'll have "n/4" moles of Cl2 produced.
Lastly, we need to convert moles of Cl2 to volume using the ideal gas law:
V = nRT / P
V = (n/4) x (0.0821 L·atm/(mol·K) x 273.15 K) / (1 atm)
By substituting the value of "n" that we calculated earlier, we can calculate the volume of Cl2 produced in liters.
It's important to note that this calculation assumes the reaction goes to completion and that all gases behave ideally under STP conditions.
Write the equation and balance it.
You can take a shortcut and use L as mols in an all gaseous equation.
Use the coefficients in the balanced equation to convert L HCl to L O2.
L O2 = L HCl x (?mol O2/?mol HCl) = ?