Oxygen gas and water are produced by the decomposition of hydrogen peroxide. If 10.0 mol of H2O2 decomposes, what volume of oxygen will be produced? Assume the density of oxygen is 1.429g/L. 2H2O2= 2H2O+O2

2H2O2 ==> 2H2O + O2

mol H2O2 = 10.0
mol O2 produced = (1/2) x mols H2O2 = 1/2 x 10 = 5 mol
At STP each mol will occupy 22.4 L.

To find the volume of oxygen produced, we need to use the balanced chemical equation and the molar ratio between H2O2 and O2.

From the balanced equation:
2 H2O2 → 2 H2O + O2

We can see that for every 2 moles of H2O2, we get 1 mole of O2. Therefore, the molar ratio is 2:1.

Given that we have 10.0 mol of H2O2, we can calculate the moles of O2 produced:

10.0 mol H2O2 × (1 mol O2 / 2 mol H2O2) = 5.0 mol O2

Now, to find the volume of oxygen, we need to use the ideal gas law equation:

PV = nRT

Where:
P = pressure in atm (Assuming the pressure remains constant)
V = volume in liters (What we need to find)
n = number of moles of the gas (5.0 mol O2)
R = gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin (Assuming room temperature)

First, we need to convert the density of oxygen from g/L to mol/L. The molar mass of oxygen (O2) is approximately 32 g/mol.

1.429 g/L × (1 mol O2 / 32 g) = 0.04465 mol/L

Now, we can rearrange the ideal gas law equation to solve for volume (V):

V = nRT / P

Assuming room temperature is around 298 K and the pressure is constant, we can substitute the values into the equation:

V = (5.0 mol O2) × (0.0821 L·atm/(mol·K)) × (298 K) / 1 atm

V ≈ 12.18 L

Therefore, approximately 12.18 liters of oxygen gas will be produced when 10.0 moles of hydrogen peroxide decompose.