Hydrogen peroxide decomposes to water and
oxygen gas by the following reaction
2H2O2(l) → 2H2O(l) + O2(g)
If 30 g of hydrogen peroxide decomposes completely, what volume of oxygen gas is formed
at STP?
1. 19.77 L
2. 22.41 L
3. 9.89 L
4. 10.79 L
5. 24.46 L
Well, let's whip up some chemistry and solve this question!
First, we need to determine the number of moles of hydrogen peroxide present. To do this, we divide the given mass by the molar mass of hydrogen peroxide. The molar mass of H2O2 is approximately 34 g/mol.
So, 30 g / 34 g/mol = 0.88 mol
According to the balanced equation, 2 moles of H2O2 produce 1 mole of O2. Therefore, for 0.88 moles of H2O2, we will get 0.44 moles of O2.
Now, we can use the ideal gas law to find the volume of O2 at STP (Standard Temperature and Pressure). The molar volume of a gas at STP is 22.4 L/mol.
So, 0.44 mol x 22.4 L/mol = 9.86 L
Now, let me sprinkle some humor on this calculation. If converting moles to liters were a magic trick, this answer would make a magic show proud. The correct volume of oxygen gas formed at STP is approximately 9.86 L. So, option 3, which is 9.89 L, is the closest choice. You were almost there! Keep up the good work!
To find the volume of oxygen gas formed at STP, we can use the ideal gas law equation: PV = nRT, where:
P = pressure (which is 1 atm at STP)
V = volume
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (which is 273.15 K at STP)
First, we need to determine the number of moles of oxygen gas formed. Since the balanced equation shows that 2 moles of H2O2 produce 1 mole of O2, we can find the number of moles of O2 by dividing the number of moles of H2O2 by 2.
1 mole of H2O2 is equal to 34.01 g
30 g of H2O2 is equal to (30 g / 34.01 g/mol) = 0.881 moles of H2O2
Therefore, the number of moles of O2 formed is (0.881 moles of H2O2 / 2) = 0.4405 moles of O2.
Now we can calculate the volume of the gas using the ideal gas law.
V = (nRT) / P
V = (0.4405 moles * 0.0821 L·atm/(mol·K) * 273.15 K) / 1 atm
V = 9.89 L
So, the volume of oxygen gas formed at STP is 9.89 L.
Therefore, the correct answer is option 3: 9.89 L.
To find the volume of oxygen gas formed, we need to use the stoichiometry of the reaction and the ideal gas law.
Step 1: Calculate the number of moles of hydrogen peroxide (H2O2) using its molar mass. The molar mass of H2O2 is 34.02 g/mol.
30 g H2O2 × (1 mol H2O2 / 34.02 g H2O2) = 0.882 mol H2O2
Step 2: Use the stoichiometry of the reaction to find the number of moles of oxygen gas (O2) produced. From the balanced equation, we see that the ratio of moles of H2O2 to moles of O2 is 2:1.
0.882 mol H2O2 × (1 mol O2 / 2 mol H2O2) = 0.441 mol O2
Step 3: Convert moles of O2 to volume at STP using the ideal gas law. At STP (standard temperature and pressure), 1 mole of any ideal gas occupies 22.41 L.
0.441 mol O2 × 22.41 L/mol = 9.89 L
The volume of oxygen gas formed at STP is therefore 9.89 L.
So, the correct answer is option 3: 9.89 L.
2H2O2(l) → 2H2O(l) + O2(g)
How many moles H2O2 in 30 g? That's moles = g/molar mass = 30/34 = ?
Look at the balanced equation. 2 mols H2O2 gives 1 mol O2; therefore ? moles H2O2 will produce ?/2 moles O2 gas.
You know that 1 mole O2 gas @ STP occupies 22.4 L. So what volume will ?/2 moles occupy?
Post your work if you get stuck.