Question

How many grams of KO₂ are needed to produce 660.0L of O2 at 19.0C and 1.00atm?

4KO₂(s) + 2CO₂(g) --> 2K₂CO₃(s) + 3O₂(g)

Answer 1

To produce 660.0L of O2, 4812.154 grams of KO₂ are needed.To find the mass of KO₂ needed to produce 660.0 L of O₂ at 19.0°C and 1.00 atm, apply the ideal gas law to calculate the moles of O₂, and then use stoichiometry and the molar mass of KO₂ to determine the required mass.

Step-by-step explanation:

To determine the amount of KO₂ needed to produce 660.0L of O2, we first need to calculate the number of moles of O2 required. Since the balanced equation shows that 4 moles of KO₂ produce 3 moles of O2, we can set up a proportion to solve for the moles of O2. Then, we can use the molar mass of KO₂ to convert moles of O2 to grams of KO₂. Finally, we can convert grams of KO₂ to the desired units, which are needed to produce 660.0L of O2.

Let's start by calculating the moles of O2:

660.0L of O2 x (1 mole O2 / 22.4L of O2) = 29.464 moles of O2

Next, let's use the balanced equation to set up a proportion:

4 moles of KO₂ / 3 moles of O2 = x moles of KO₂ / 29.464 moles of O2

Solving for x, we find that x = 39.286 moles of KO₂.

Finally, let's convert moles of KO₂ to grams:

39.286 moles of KO₂ x (122.55 g of KO₂ / 1 mole of KO₂) = 4812.154 g of KO₂

Therefore, 4812.154 grams of KO₂ are needed to produce 660.0L of O2.