Question

Use the balanced equation to solve the stoichiometry problems below.

(a) What mass of KO2 produces 235 g of O2?

(A) 122.5 g
(B) 235 g
(C) 470 g
(D) 352.5 g

(b) How many moles of O2 can be removed by 12.3 L of KO2?

(A) 0.25 moles
(B) 0.5 moles
(C) 1 mole
(D) 2 moles

Answer 1

In part (a), the mass of KO2 that produces 235 g of O2 is 122.5 g. In part (b), 12.3 L of KO2 can remove 5.16 moles of O2.

Step-by-step explanation:

In order to solve stoichiometry problems using a balanced equation, we need to convert between mass and moles. Let's start with part (a). The balanced equation for the reaction is 4KO2(s) + 2CO2(g) → 2K2CO3(s) + 3O2(g). From the equation, we can see that for every 4 moles of KO2, 3 moles of O2 are produced. We can set up a proportion to find the mass of KO2 that produces 235 g of O2:

(x g KO2) / (235 g O2) = (4 mol KO2) / (3 mol O2)

Solving for x gives us x = 122.5 g. Therefore, the answer to part (a) is 122.5 g. For part (b), we can use the ideal gas law to solve. The molar volume of a gas at standard temperature and pressure (STP) is 22.4 L/mol. The balanced equation shows that 4 moles of KO2 produce 3 moles of O2. So, for every 6.88 moles of KO2, the number of moles of O2 can be found using the ratio:

(x mol O2) / (6.88 mol KO2) = (3 mol O2) / (4 mol KO2)

Solving for x gives us x = 5.16 moles. Therefore, the answer to part (b) is 5.16 moles.