Question

Potassium nitrate decomposes to potassium nitrite and oxygen. What volume, in L, of O2 can be produced from the decomposition of 50.0 g of KNO3 at 35 °C and 1.19 atm?

a. 10.5 L
b. 5.25 L
c. 24.9 L
d. 0.596 L

Answer 1

The volume of oxygen gas produced from the decomposition of potassium nitrate can be calculated by using the appropriate conversion factors and the given mass of potassium nitrate. The correct volume of oxygen gas is approximately 0.745 L.

Step-by-step explanation:

To calculate the volume of oxygen gas produced from the decomposition of potassium nitrate, we first need to convert the mass of potassium nitrate to moles. The molar mass of KNO3 is 101.1 g/mol, so 50.0 g of KNO3 is equal to 0.494 mol.

The balanced chemical equation for the decomposition of KNO3 is: 2KNO3 → 2KNO2 + 3O2

From the equation, we can see that for every 2 moles of KNO3 decomposed, 3 moles of O2 are produced. Therefore, we can set up a ratio:

3 mol O2 / 2 mol KNO3 = x L O2 / 0.494 mol KNO3

Solving for x, we find that the volume of O2 produced is approximately 0.745 L. Therefore, the correct option is (d) 0.745 L.