2H2O2 ==> 2H2O + O2
mols O2 gas = 0.460L/22.4 = ?
Using the coefficients in the balanced equation, convert mols O2 gas to mols H2O2.
Now convert mols H2O2 to grams. g = mols x molar mass.
mols O2 gas = 0.460L/22.4 = ?
Using the coefficients in the balanced equation, convert mols O2 gas to mols H2O2.
Now convert mols H2O2 to grams. g = mols x molar mass.
The balanced equation for the decomposition of hydrogen peroxide is:
2H2O2 -> 2H2O + O2
According to the equation, for every 2 moles of hydrogen peroxide, we get 1 mole of oxygen gas.
1 mole of any gas at STP occupies 22.4 liters. Since we have 0.460L of oxygen gas, we can calculate the number of moles using the ideal gas law:
PV = nRT
Where:
P = pressure (STP is 1 atm)
V = volume (0.460L)
n = number of moles (unknown)
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (273.15 K)
Rearranging the equation, we get:
n = PV / RT
Plugging in the values:
n = (1 atm) * (0.460 L) / (0.0821 L·atm/(mol·K)) * (273.15 K)
Calculating n, we find:
n ≈ 0.0177 mol
Now, using the stoichiometry of the reaction, we know that for every 2 moles of hydrogen peroxide, we get 1 mole of oxygen gas. Therefore, we need to multiply the number of moles of oxygen gas by a factor of 2:
Moles of hydrogen peroxide = 0.0177 mol * (2 mol of H2O2 / 1 mol of O2)
Calculating the moles of hydrogen peroxide, we find:
Moles of hydrogen peroxide = 0.0354 mol
Next, we need to convert moles to grams. The molar mass of hydrogen peroxide (H2O2) is 34.0147 g/mol.
Mass of hydrogen peroxide = Moles of hydrogen peroxide * Molar mass of H2O2
Calculating the mass, we find:
Mass of hydrogen peroxide ≈ 0.0354 mol * 34.0147 g/mol
Mass of hydrogen peroxide ≈ 1.204 g
Therefore, approximately 1.204 grams of hydrogen peroxide would be needed to obtain 0.460L of oxygen gas at STP.
The balanced chemical equation for the decomposition of hydrogen peroxide is:
2 H2O2 → 2 H2O + O2
From this equation, we can see that for every 2 moles of hydrogen peroxide (H2O2) decomposed, we obtain 1 mole of oxygen gas (O2).
First, we calculate the number of moles of oxygen gas by using the ideal gas law:
PV = nRT
Where:
P = pressure (at STP, it is 1 atm)
V = volume (0.460 L)
n = number of moles
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature (at STP, it is 273 K)
n = PV / RT
= (1 atm) x (0.460 L) / (0.0821 L.atm/mol.K) x (273 K)
≈ 0.0208 mol
Next, we use the stoichiometry of the balanced equation to determine the moles of hydrogen peroxide needed. From the equation, we know that 2 moles of hydrogen peroxide produce 1 mole of oxygen gas.
Therefore, 0.0208 mol of oxygen gas corresponds to (0.0208 mol) x (2 mol H2O2 / 1 mol O2) = 0.0416 mol of hydrogen peroxide.
Lastly, we need to calculate the mass of hydrogen peroxide using its molar mass, which is 34.02 g/mol.
Mass = moles x molar mass
= (0.0416 mol) x (34.02 g/mol)
≈ 1.42 g
Therefore, approximately 1.42 grams of hydrogen peroxide is needed to obtain 0.460 L of oxygen gas at STP.