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 react 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 equation for the reaction 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. We can use this ratio to calculate the number of moles of Cl2 produced in the reaction.
First, we need to convert the given volume of HCl gas to its corresponding number of moles. To do this, we can use the ideal gas law equation, PV = nRT.
At STP, the temperature (T) is 273.15 K and the pressure (P) is 1 atmosphere. Therefore, we can calculate the number of moles of HCl as follows:
n(HCl) = (PV) / (RT)
= (1 atm * 0.98 L) / (0.0821 L·atm/(mol·K) * 273.15 K)
≈ 0.0457 moles of HCl
Now that we have the number of moles of HCl, we can use the mole ratio from the balanced equation to determine the number of moles of Cl2 produced.
From the stoichiometry, 4 moles of HCl produce 2 moles of Cl2. Therefore, we have:
n(Cl2) = (0.0457 moles HCl) * (2 moles Cl2 / 4 moles HCl)
≈ 0.02285 moles of Cl2
Finally, we can convert the number of moles of Cl2 to liters by using the ideal gas law equation in reverse:
V(Cl2) = n(Cl2) * (RT) / P
= (0.02285 moles * 0.0821 L·atm/(mol·K) * 273.15 K) / (1 atm)
≈ 0.521 L
Therefore, approximately 0.521 liters of chlorine gas can be produced when 0.98 L of HCl reacts with excess O2 at STP.