The reaction between potassium superoxide, KO2, and CO2 4 KO2+2 CO2 ⟶2 K2 CO3+3 O2 is used as a

The reaction between potassium superoxide, KO2, and CO2 4...

The reaction between potassium superoxide, $\mathrm{KO}_{2},$ and $\mathrm{CO}_{2}$ $$ 4 \mathrm{KO}_{2}+2 \mathrm{CO}_{2} \longrightarrow 2 \mathrm{K}_{2} \mathrm{CO}_{3}+3 \mathrm{O}_{2} $$ is used as a source of $\mathrm{O}_{2}$ and absorber of $\mathrm{CO}_{2}$ in self-contained breathing equipment used by rescue workers. (a) How many moles of $\mathrm{O}_{2}$ are produced when 0.400 $\mathrm{mol}$ of $\mathrm{KO}_{2}$ reacts in this fashion? (b) How many grams of $\mathrm{KO}_{2}$ are needed to form 7.50 $\mathrm{g}$ of $\mathrm{O}_{2} ?$ (c) How many grams of $\mathrm{CO}_{2}$ are used when 7.50 $\mathrm{g}$ of $\mathrm{O}_{2}$ are produced?

The reaction between potassium superoxide, $\mathrm{KO}_{2},$ and $\mathrm{CO}_{2}$
$$
4 \mathrm{KO}_{2}+2 \mathrm{CO}_{2} \longrightarrow 2 \mathrm{K}_{2} \mathrm{CO}_{3}+3 \mathrm{O}_{2}
$$
is used as a source of $\mathrm{O}_{2}$ and absorber of $\mathrm{CO}_{2}$ in self-contained breathing equipment used by rescue workers.
(a) How many moles of $\mathrm{O}_{2}$ are produced when 0.400 $\mathrm{mol}$ of $\mathrm{KO}_{2}$ reacts in this fashion?
(b) How many grams of $\mathrm{KO}_{2}$ are needed to form 7.50 $\mathrm{g}$ of $\mathrm{O}_{2} ?$
(c) How many grams of $\mathrm{CO}_{2}$ are used when 7.50 $\mathrm{g}$ of $\mathrm{O}_{2}$ are produced?

Key Concepts

Stoichiometry

Stoichiometry is the study of the quantitative relationships between the reactants and products in a chemical reaction. It involves using the coefficients from a balanced chemical equation to determine the proportions in which substances react and are formed. This concept is fundamental for calculating how much of each substance is needed or produced in a chemical reaction.

Balanced Chemical Equations

Balanced chemical equations represent reactions by ensuring that the number of atoms for each element is the same on both sides of the equation. This balance is achieved by adjusting the coefficients, which are essential for determining the molar relationships between reactants and products. It serves as the foundation for all stoichiometric calculations.

Moles-to-Grams Conversion

Moles-to-grams conversion is a method used to convert the amount of a substance from moles, a measure of quantity, to grams, a measure of mass, using the substance's molar mass. This conversion is crucial when performing stoichiometric calculations as it links the theoretical mole ratios from the balanced equation to practical laboratory measurements.

Transcript

00:01 So in this problem, we are given this reaction equation between k -o -2 and co2, where k -2, co3, and o2 or oxygen is produced.

00:13 So the first question here is, how many moles of o2 are produced when 0 .4 mole of k -o -2 reacts in this reaction? so from the balanced chemical equation, we can see that 4 -mull of, k -o -2 will produce 3 mole of oxygen.

00:41 That means from here, we can find out how many mole of o2 will be produced from 0 .4 -mull of k -o -2.

00:51 That means 0 .4 -4 -0 -2 will produce 3 over 4 times 0 .4 mall, which is equal to 0 .3 mole or 2.

01:23 That means 0 .4 mol of ko2 will produce 0 .3 mole of auto.

01:30 Now in the second part of this problem, we are asked to find out how many grams of ko2 are needed to form 7 .5 gram of o2.

01:39 So what we have to do is we just have to convert this gram into number of moles.

01:45 And again, we can just find out how many moles of co2 will be required to produce that much more of o2 and from there we can find out how many grams of co2 will be needed to form 7 .5 gram o2 so again here we can see that so first let's find out the number of moles of o2 for 7 .5 gram or 2 so as you know that the number of moles is basically the math over molar mass that means 7 .5 gram o2 is 7 .5 over 32 which is the molar mass of o two since for one oxygen it is 16 the molar mass of a single oxygen atom is 16 that means for o two molecule it is 32 so if we divide 7 .5 by 32 we will get a value of 0 .234 so this is the number of mole of 7 .5 gram o2 now we have to find out how many moles of co2 we need to produce 0 .234 mol of o2 so since um three mole o two is produced from formal k o2 is produced from four mall k o2 that means 0 .2 will be produced from 4 by 3 times 0 .2...

Please give Ace some feedback