how many liters of chlorine gas can be produced when .98L of HCl react with excess O at STP? 4HCl(g) + O2(g) => 2Cl(g) + 2H2O(g)
for each HCl liter, you get 1/2 liter of Cl2. Check that ratio.
To determine the number of liters of chlorine gas that can be produced when 0.98L of HCl reacts with excess O2 at STP, we first need to calculate the number of moles of HCl using the ideal gas law equation:
PV = nRT
Where:
P = pressure (STP = 1 atm)
V = volume of gas (0.98L)
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (STP = 273.15K)
Rearranging the equation to solve for n:
n = PV / RT
n = (1 atm) * (0.98L) / (0.0821 L·atm/mol·K * 273.15K)
n ≈ 0.0356 moles of HCl
Next, we can use the balanced equation to determine the stoichiometry between HCl and Cl2. According to the balanced equation:
4HCl(g) + O2(g) → 2Cl2(g) + 2H2O(g)
We can see that 4 moles of HCl produce 2 moles of Cl2. Therefore, if 0.0356 moles of HCl react completely, it will produce:
(0.0356 mol HCl) * (2 mol Cl2 / 4 mol HCl) = 0.0178 moles of Cl2
Finally, since we know that 1 mole of any ideal gas at STP occupies 22.4L of volume, we can calculate the volume of Cl2 produced:
(0.0178 mol Cl2) * (22.4 L/mol) = 0.398 L of Cl2
Therefore, when 0.98L of HCl reacts with excess O2 at STP, approximately 0.398L of chlorine gas can be produced.