Hydrogen peroxide(H2O2) slowly decomposes to form water and oxygen gas. Calculate the mass of Hydrogen peroxide needed to obtain 0.460L of oxygen gas at STP

To calculate the mass of hydrogen peroxide needed to obtain 0.460L of oxygen gas at STP, we need to use the stoichiometry of the reaction and the ideal gas law.

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.

To calculate the mass of hydrogen peroxide needed to obtain 0.460 L of oxygen gas at STP, we need to apply the concept of stoichiometry.

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.

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.