if 8.6L of H2 reacted with 4.3L of O2 at STP, what is the volume of gaseous water collected (assuming none of it condenses)? 2H2(g) + O2(g) => 2H2O(g)
This may be a limiting reagent problem. You must determine which is the limiting reagent. To do this, I usually simply calculate how much of one reagent is needed if we start with the other. For example,
8.6 L H2 will require how much O2?
8.6 x (1 mol O2/2 mols H2) = 4.3 and we have exactly 4.3; therefore, both are limiting and we have exactly the amount of each we need. Use this same procedure to determine the amount of H2O produced.
To determine the volume of gaseous water collected, we can use the balanced equation and the ideal gas law to solve for the volume.
Step 1: Write and balance the chemical equation.
2H2(g) + O2(g) → 2H2O(g)
Step 2: Determine the moles of hydrogen gas (H2) and oxygen gas (O2) based on the given volumes.
Using the ideal gas law equation, PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature. Since the reaction is at STP (standard temperature and pressure), we can use the values:
P = 1 atm
T = 273.15 K
R = 0.0821 L·atm/(mol·K)
For hydrogen gas (H2):
n(H2) = (V(H2) * P) / (R * T)
n(H2) = (8.6 L * 1 atm) / (0.0821 L·atm/(mol·K) * 273.15 K)
n(H2) = 0.3679 mol
For oxygen gas (O2):
n(O2) = (V(O2) * P) / (R * T)
n(O2) = (4.3 L * 1 atm) / (0.0821 L·atm/(mol·K) * 273.15 K)
n(O2) = 0.1845 mol
Step 3: Determine the limiting reactant.
Compare the moles of each reactant to determine which one is limiting. The reactant that produces fewer moles of product is the limiting reactant.
Since the stoichiometric ratio of H2 to H2O is 2:2, the moles of water produced are equal to the moles of hydrogen gas: n(H2O) = 0.3679 mol.
Step 4: Determine the volume of gaseous water produced.
Using the ideal gas law, we can calculate the volume of gaseous water:
V(H2O) = n(H2O) * (R * T) / P
V(H2O) = 0.3679 mol * (0.0821 L·atm/(mol·K) * 273.15 K) / 1 atm
V(H2O) = 7.13 L
Therefore, the volume of gaseous water collected is 7.13 liters.
To determine the volume of gaseous water collected, we need to use the balanced equation to determine the stoichiometry of the reaction.
The balanced equation is:
2H2(g) + O2(g) => 2H2O(g)
From the equation, we can see that the ratio of H2 to H2O is 2:2, or 1:1. This means that for every 1 mole of H2 consumed, 1 mole of H2O is produced.
First, we need to convert the given volumes of H2 and O2 to moles using the ideal gas law:
PV = nRT
At STP (standard temperature and pressure):
T = 273 K (0 degrees Celsius)
P = 1 atm
For H2:
V(H2) = 8.6 L
n(H2) = (P x V) / (R x T) = (1 atm x 8.6 L) / (0.0821 L.atm/mol.K x 273 K) = 0.404 mol
For O2:
V(O2) = 4.3 L
n(O2) = (P x V) / (R x T) = (1 atm x 4.3 L) / (0.0821 L.atm/mol.K x 273 K) = 0.204 mol
Now, since the stoichiometry of the reaction tells us that 2 moles of H2 produces 2 moles of H2O, we can determine the number of moles of H2O produced.
n(H2O) = n(H2) = 0.404 mol
Finally, we can convert the moles of gaseous water to volume using the ideal gas law:
V(H2O) = (n x R x T) / P = (0.404 mol x 0.0821 L.atm/mol.K x 273 K) / 1 atm ≈ 8.8 L
Therefore, assuming none of the water vapor condenses, the volume of gaseous water collected is approximately 8.8 liters.