Question

If 10 grams of potassium chlorate decompose to form potassium chloride and oxygen gas inside a 500 mL container at a temperature of 45 °C. what will the pressure be inside the container?​

Answer 1

Answer:

10 atm

Explanation:

There's a lot to do here, but lets take it one step at a time. First, let's write a balanced equation for the decomposition of potassium chlorate into potassium chloride and oxgyen gas.

2 KClO3 → 2 KCl + 3 O2

Now let's find the moles of the KClO3 (molar mass 122.55 g/mol) that we have take 10 g/122.55 g/mol, grams will cancel and we are left with 0.0816 moles. lets divide that by two since we have a two in front of the KClO3 in the equation, and then multiply that number by 5 since it's the total moles of products, in summary, multiply by 5/2 to get 0.204 moles.

Now that we know the moles of our products, let's plug some stuff into the ideal gas law PV = nRT. We are looking for P so let's solve for that. P = (nRT)/V, now let's plug in our values. Make sure V is converted to liters so 0.5 L. And convert celcius to kelvin by adding 273

P = ((0.204 moles)(318 K)(0.08206 L atm mol^-1 K^-1))/0.5 L

A lot of units cancel, and we get about 10.65 atm, if you don't want the answer in atm, you can find a conversion equation. But let's round to sig figs for now, which will bring us to 10 atm.