If 100 mL of hydrogen peroxide decomposes how many liters of O2 is formed

To find the number of liters of O2 formed when 100 mL of hydrogen peroxide decomposes, we need to know the stoichiometry of the reaction. The balanced equation for the decomposition of hydrogen peroxide is:

2 H2O2 (l) -> 2 H2O (l) + O2 (g)

From the balanced equation, we can see that two moles of hydrogen peroxide produce one mole of oxygen gas.

To calculate the number of moles of hydrogen peroxide in 100 mL, we need to know the molarity of the hydrogen peroxide solution. Let's assume it is 1 M (moles per liter).

Molarity (M) is defined as moles of solute divided by the volume of the solution in liters. In this case, the volume is given as 100 mL, so we need to convert it to liters by dividing by 1000:

100 mL / 1000 mL/L = 0.1 L

Now, we can calculate the number of moles of hydrogen peroxide:

moles of H2O2 = Molarity * Volume
= 1 M * 0.1 L
= 0.1 moles

From the balanced equation, we know that two moles of hydrogen peroxide produce one mole of oxygen gas. So, if 0.1 moles of hydrogen peroxide decompose, we would expect the formation of half that amount of moles of oxygen gas:

moles of O2 = 0.1 moles / 2
= 0.05 moles

Finally, we can convert moles of oxygen gas to liters using the ideal gas law, assuming standard temperature and pressure (STP). At STP, one mole of any ideal gas occupies 22.4 liters. Therefore:

liters of O2 = moles of O2 * 22.4 L/mol
= 0.05 moles * 22.4 L/mol
= 1.12 liters

So, when 100 mL of hydrogen peroxide decomposes at STP, 1.12 liters of oxygen gas is formed.