What mass of water, in grams, would have to be decomposed to produce 26.3 L of molecular oxygen at STP?
Since 22.4 liters of a gas is 1 mole at STP, you would need 26.3/22.4 = 1.17 moles of O2.
I assume you are talking about the electrolysis reaction
2 H2O -> 2H2 + O2
Clearly, you need twice as many moles of water, or 2.34 moles, which would have a mass of 42.1 g.
To determine the mass of water required to produce 26.3 liters of molecular oxygen at STP, we need to use the concept of stoichiometry and the molar volume of gases at STP.
1. Determine the balanced chemical equation showing the decomposition of water into molecular oxygen:
2H2O -> 2H2 + O2
2. Calculate the molar volume of a gas at STP, which is 22.4 L/mol.
3. Determine the stoichiometric ratio between water and molecular oxygen. From the balanced equation, we can see that 2 moles of water produce 1 mole of molecular oxygen.
4. Convert the given volume of molecular oxygen to moles. Since the molar volume of gases at STP is 22.4 L/mol, we have:
26.3 L O2 * (1 mol O2 / 22.4 L O2) = 1.17 mol O2
5. Calculate the moles of water needed based on the stoichiometric ratio:
1.17 mol O2 * (2 mol H2O / 1 mol O2) = 2.34 mol H2O
6. Calculate the molar mass of water (H2O), which is approximately 18 g/mol.
7. Finally, calculate the mass of water needed:
2.34 mol H2O * (18 g H2O / 1 mol H2O) = 42.12 g
Therefore, approximately 42.12 grams of water would have to be decomposed to produce 26.3 L of molecular oxygen at STP.
To find the mass of water that needs to be decomposed to produce 26.3 L of molecular oxygen at STP, you first need to determine the number of moles of oxygen gas. Once you have that information, you can use the molar ratio between water and oxygen to find the mass of water.
Step 1: Determine the number of moles of oxygen gas.
To do this, you can use the ideal gas law, which states PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin (STP is 273.15 K).
In this case, you know that the volume of oxygen gas is 26.3 L at STP, so you can plug in the values:
PV = nRT
(1 atm) (26.3 L) = n (0.0821 L·atm/mol·K) (273.15 K)
Solving for n, the number of moles of oxygen gas:
n = (1 atm) (26.3 L) / (0.0821 L·atm/mol·K) (273.15 K)
Step 2: Use the molar ratio to find the mass of water.
The balanced equation for the decomposition of water is:
2H2O(l) -> 2H2(g) + O2(g)
From the equation, you can see that for every 2 moles of water, you get 1 mole of O2. Therefore, the molar ratio of water to oxygen gas is 2:1.
Since you now know the number of moles of oxygen gas from Step 1, you can use the molar ratio to find the moles of water needed, and then convert that to grams.
Assuming the molar mass of water is 18.015 g/mol (H2O), you can calculate the mass of water:
mass of water = moles of water × molar mass of water
Substituting the values:
mass of water = (n/2) × (18.015 g/mol)
Now, plug in the value of n from Step 1 into the equation to find the mass of water.