How many liters of chlorine gas can be produced when 0.98 L of HCl react with excess O2 at STP?
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To determine the amount of chlorine gas that can be produced when 0.98 L of HCl reacts with excess O2 at STP (Standard Temperature and Pressure), we need to use stoichiometry and the balanced equation of the reaction.
Let's first write the balanced equation for the reaction between HCl and O2:
4 HCl + O2 → 2 Cl2 + 2 H2O
From the balanced equation, we can see that for every 4 moles of HCl, we produce 2 moles of Cl2 gas.
To solve this problem, follow these steps:
Step 1: Convert the given volume of HCl gas to moles. We can use the ideal gas law equation, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. Since the conditions are at STP, we can use the molar volume of an ideal gas at STP, which is 22.4 L/mol.
n(HCl) = V(HCl) / V(molar) = 0.98 L / 22.4 L/mol
Step 2: Calculate the number of moles of Cl2 gas produced. Since the stoichiometric ratio of Cl2 to HCl is 2:4, we can use a simple ratio:
n(Cl2) = (n(HCl) / 4) * 2
Step 3: Convert the moles of Cl2 gas to liters using the molar volume at STP:
V(Cl2) = n(Cl2) * V(molar) = n(Cl2) * 22.4 L/mol
Now you can substitute the values into the formula:
n(HCl) = 0.98 L / 22.4 L/mol = 0.04375 mol
n(Cl2) = (0.04375 mol / 4) * 2 = 0.02188 mol
V(Cl2) = 0.02188 mol * 22.4 L/mol = 0.4892 L
Therefore, when 0.98 L of HCl reacts with excess O2 at STP, approximately 0.4892 liters of Cl2 gas can be produced.