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)
When all are gases one can use L as if they were moles.
8.6L H2 x (2 moles H2O/2 moles H2) = 8.6*1 = 8.6L
4.3L O2 x (2 moles H2O/1 mole O2) = 4.3*2 = 8.6.
This is a limiting reagent problem in which NEITHER is limiting; i.e., you will form 8.6 L H2O as a gas.
Oh, the classic reaction of hydrogen and oxygen to form water! Your question has me all excited, so let's dive into the calculations with a splash of humor.
Firstly, let's find the limiting reactant. We'll need equal amounts of H2 and O2 to react fully, so if one is in excess, it's like trying to find a clown in a car filled with other clowns – it's just not possible.
Given that the ratio between H2 and O2 is 2:1, we can assume that the O2 will be the limiting reactant since we only have half the amount we need. So, bye-bye, O2!
Now let's figure out the number of moles of H2 and H2O formed. Using the gas laws, we know that at STP, 1 mole of any gas occupies 22.4 liters of volume. Since we have 8.6 liters of H2, we can calculate:
8.6 L H2 * (1 mol H2 / 22.4 L H2) = ??? moles of H2
Using stoichiometry, we find that for every 2 moles of H2, we get 2 moles of H2O. So, the number of moles of H2O is the same as the number of moles of H2:
??? moles of H2O = ??? moles of H2
Finally, to find the volume of gaseous water formed, we'll use the gas law again. At STP, 1 mole of any gas occupies 22.4 liters of volume.
??? moles of H2O * (22.4 L H2O / 1 mol H2O) = ??? liters of gaseous water
And there you have it – the magical volume of gaseous water collected! I hope my shenanigans helped you navigate through those calculations. Remember, in the world of chemistry, humor always makes the reactions more bubbly!
To find the volume of gaseous water collected, we need to use the balanced equation to determine the stoichiometry of the reaction. According to the balanced equation:
2H2(g) + O2(g) => 2H2O(g)
We can see that 2 moles of H2 react with 1 mole of O2 to produce 2 moles of H2O.
First, let's calculate the number of moles of H2 and O2 present using the ideal gas law equation:
n = PV / RT
where:
n = number of moles
P = pressure (which is STP, so P = 1 atm)
V = volume of gas in liters
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature in Kelvin (which is STP, so T = 273 K)
For H2:
n(H2) = (1 atm * 8.6 L) / (0.0821 L·atm/mol·K * 273 K) = 0.356 moles of H2
For O2:
n(O2) = (1 atm * 4.3 L) / (0.0821 L·atm/mol·K * 273 K) = 0.178 moles of O2
Next, we need to determine the limiting reactant, which is the reactant that is completely consumed. The limiting reactant can be determined by comparing the ratio of moles of reactants to their coefficients in the balanced equation.
From the balanced equation, we see that 2 moles of H2 react with 1 mole of O2. Therefore, for the given number of moles, we need twice as many moles of H2 as O2 for complete reaction.
Since we have 0.356 moles of H2 and only 0.178 moles of O2, O2 is the limiting reactant. This means that O2 will be completely consumed, and the amount of H2 will not be a limiting factor.
Now, we can determine the number of moles of H2O produced using the stoichiometry from the balanced equation. Since the reaction produces 2 moles of H2O for every 1 mole of O2, we can calculate:
n(H2O) = 2 * n(O2) = 2 * 0.178 = 0.356 moles of H2O
Finally, we can find the volume of gaseous water collected using the ideal gas law:
V(H2O) = n(H2O) * RT / P
where:
V(H2O) = volume of gaseous water in liters
n(H2O) = number of moles of H2O
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature in Kelvin (which is STP, so T = 273 K)
P = pressure (which is also STP, so P = 1 atm)
Substituting the values:
V(H2O) = (0.356 mol * 0.0821 L·atm/mol·K * 273 K) / 1 atm = 8.72 L
Therefore, the volume of gaseous water collected is 8.72 liters.