Question

Potassium superoxide, KO2, reacts with carbon dioxide to form potassium carbonate and oxygen: This reaction makes potassium superoxide useful in a self-contained breathing apparatus. How much O2 could be produced from 2.50 g of KO2 and 4.50 g of CO2?4 KO2(s) + 2 CO2(g) → 2 K2CO3(s) + 3 O2(g)

Answer 1

Answer:

There will be produced 0.8448 grams O2

Explanation:

Step 1: Data given

Mass of KO2 = 2.50 grams

Mass of CO2 = 4.50 grams

Molar mass KO2 = 71.1 g/mol

Molar mass of CO2 = 44.01 g/mol

Molar mass of O2 = 32 g/mol

Step 2: The balanced equation

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

Step 3: Calculate moles KO2

Moles KO2 = mass KO2 / Molar mass KO2

Moles KO2 = 2.50 grams / 71.1 g/mol

Moles KO2 = 0.0352 moles

Step 4: Calculate moles of CO2:

Moles CO2 = mass CO2 / Molar mass CO2

Moles CO2 = 4.50 grams / 44.01 g/mol

Moles CO2 = 0.102 moles

Step 5: calculate limiting reactant

For 4 moles of KO2 consumed, we need 2 moles of CO2 to produce 2 moles of K2CO3 and 3 moles of O2

KO2 is the limiting reactant. It will completely be consumed. (0.0352 moles).

CO2 is in excess. There will react 0.0352/2 = 0.0176 moles of CO2

There will remain 0.102 - 0.0176 = 0.0844 moles CO2

Step 6: Calculate moles O2 produced

For 4 moles of KO2 consumed, we need 2 moles of CO2 to produce 2 moles of K2CO3 and 3 moles of O2

For 0.0352 moles KO2 consumed, we have 3/4 * 0.0352 = 0.0264 moles of O2 produced

Step 7: Calculate mass of 02 produced

Mass O2 produced = Moles O2 * Molar mass 02

Mass O2 produced = 0.0264 moles * 32 g/mol

Mass O2 produced = 0.8448 grams O2

There will be produced 0.8448 grams O2

Answer 2

Answer:

0.8448 grams of oxygen gas is produced in the reaction of potassium superoxide with carbon dioxide.

Explanation:

The reaction of Potassium superoxide $\rm (KO_2)$ reacts with Carbon dioxide $\rm (CO_2)$ to form Potassium carbonate $\rm (K_2CO_3)$ and Oxygen $\rm (O_2)$.

The reaction will be:

$\rm 4\;KO_2\;+2\;CO_2\;\rightarrow\;2\;K_2CO_3\;+\;3\;O_2$

4 moles of $\rm KO_2$ and 2 moles of $\rm CO_2$ is required to produce 3 moles of oxygen.

For the of 2.50 g $\rm (KO_2)$:

The moles of $\rm (KO_2)$ is, $\rm \frac{weight}{molecular weight}$

= $\rm \frac{2.50}{71.1}$

0.0352 moles of $\rm (KO_2)$ is there.

The moles of $\rm CO_2$ will be:

= $\rm \frac{4.50}{44.01}$

=0.102 moles of $\rm CO_2$ is there.

Since $\rm CO_2$ is in excess, the half of given moles of $\rm KO_2$ will be consumed.

$\rm KO_2$ consumed will be 0.0176 moles.

The left amount of $\rm CO_2$ will be = 0.102 - 0.0176

= 0.0844 moles.

4 moles of $\rm KO_2$ produces 3 moles of $\rm O_2$.

so, 0.0352 moles = 0.0264 moles of $\rm O_2$/.

The mass of $\rm O_2$ produces will be: moles $\times$ molecular weight

= 0.0264 $\times$ 32

= 0.8448 grams

From 2.50 grams of $\rm KO_2$ and 4.50 grams of $\rm CO_2$, 0.8448 grams of $\rm O_2$ has been produced.

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