The decomposition of potassium chlorite yields oxygen gas and potassium chloride. If the yield is 95%, how many grams of KClO3 are needed to produce10.0 L of O2?

Here is a worked example using KClO3 to produce oxygen. This problem requires that you work it "backwards" that is, Convert 10 L O2 to moles (moles = L/22.4), then divide by 0.95 to see how much MORE you need to produce to account for the 95% yield, then use the steps outlined in the example.

http://www.jiskha.com/science/chemistry/stoichiometry.html

To find the number of grams of KClO3 needed to produce 10.0 L of O2, we need to use stoichiometry. Stoichiometry is a way to relate the amounts of different substances in a chemical reaction.

First, we need to write the balanced chemical equation for the decomposition of potassium chlorite:

2KClO3 -> 2KCl + 3O2

From the equation, we can see that for every 2 moles of KClO3, we get 3 moles of O2.

Next, we need to convert the volume of O2 to moles. We can use the ideal gas law equation, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. Since we have the volume (10.0 L) and we assume the pressure and temperature are constant, we can rearrange the equation to solve for n:

n = PV/RT

Assuming the pressure is 1 atm and the temperature is 298 K, the ideal gas constant R is approximately 0.0821 L·atm/(mol·K).

n = (1 atm)(10.0 L) / (0.0821 L·atm/(mol·K) * 298 K)

n ≈ 0.406 moles of O2

Now, we can use the stoichiometry relationship to determine the number of moles of KClO3 needed to produce 0.406 moles of O2.

From the balanced equation, we know that for every 3 moles of O2, we need 2 moles of KClO3.

moles of KClO3 = (2 moles KClO3 / 3 moles O2) * 0.406 moles O2

moles of KClO3 ≈ 0.271 moles

Finally, we can use the molar mass of KClO3 to convert moles to grams. The molar mass of KClO3 is approximately 122.55 g/mol.

grams of KClO3 = moles of KClO3 * molar mass of KClO3

grams of KClO3 ≈ 0.271 moles * 122.55 g/mol

grams of KClO3 ≈ 33.19 grams

Therefore, approximately 33.19 grams of KClO3 are needed to produce 10.0 L of O2 with a yield of 95%.