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Question52 The reaction between potassium superoxide, KO2\mathrm{KO}_{2}, and CO2\mathrm{CO}_{2} 4KO2+2CO2⟶2 K2CO3+3O24 \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 O2\mathrm{O}_{2} and absorber of CO2\mathrm{CO}_{2} in self-contained breathing equipment used by rescue workers. (a) How many moles of O2\mathrm{O}_{2} are produced when 0.400 mol of KO2\mathrm{KO}_{2} reacts in this fashion? (b) How many grams of KO2\mathrm{KO}_{2} are needed to form 7.50 g of O2\mathrm{O}_{2} ?

Studdy Solution

STEP 1

What is this asking? Given a chemical reaction, we need to figure out how much oxygen is made from a certain amount of potassium superoxide and how much potassium superoxide is needed to make a certain amount of oxygen. Watch out! Make sure to keep track of the units (moles and grams) and use the correct molar masses!
Also, remember the importance of balanced chemical equations.

STEP 2

1. Moles of Oxygen Produced
2. Grams of Potassium Superoxide Needed

STEP 3

Alright, let's **start** with the balanced chemical equation: 4KO2+2CO2⟶2 K2CO3+3O24 \mathrm{KO}_{2} + 2 \mathrm{CO}_{2} \longrightarrow 2 \mathrm{~K}_{2}\mathrm{CO}_{3} + 3 \mathrm{O}_{2} This tells us that **4 moles** of KO2\mathrm{KO}_{2} produce **3 moles** of O2\mathrm{O}_{2}.
This is our **key** to converting between the two!

STEP 4

We're given that we have **0.400 moles** of KO2\mathrm{KO}_{2}.
To figure out how many moles of O2\mathrm{O}_{2} are produced, we can set up a proportion: 3 moles O24 moles KO2=x moles O20.400 moles KO2\frac{3 \text{ moles } \mathrm{O}_{2}}{4 \text{ moles } \mathrm{KO}_{2}} = \frac{x \text{ moles } \mathrm{O}_{2}}{0.400 \text{ moles } \mathrm{KO}_{2}}

STEP 5

Now, we just need to solve for xx!
We can **multiply** both sides by 0.4000.400 to get: x=34⋅0.400=0.300x = \frac{3}{4} \cdot 0.400 = 0.300 So, we get x=0.300x = \mathbf{0.300} **moles** of O2\mathrm{O}_{2}.
Awesome!

STEP 6

We want to produce **7.50 g** of O2\mathrm{O}_{2}.
First, we need to convert this to moles.
The molar mass of O2\mathrm{O}_{2} is **32.00 g/mol**.
So, we have: moles of O2=7.50 g32.00 g/mol=0.234375 moles\text{moles of } \mathrm{O}_{2} = \frac{7.50 \text{ g}}{32.00 \text{ g/mol}} = 0.234375 \text{ moles}

STEP 7

Now, we use the same **ratio** from before, but this time we're solving for moles of KO2\mathrm{KO}_{2}: 4 moles KO23 moles O2=y moles KO20.234375 moles O2\frac{4 \text{ moles } \mathrm{KO}_{2}}{3 \text{ moles } \mathrm{O}_{2}} = \frac{y \text{ moles } \mathrm{KO}_{2}}{0.234375 \text{ moles } \mathrm{O}_{2}}

STEP 8

Solving for yy, we **multiply** both sides by 0.2343750.234375: y=43⋅0.234375=0.3125y = \frac{4}{3} \cdot 0.234375 = 0.3125 So, we need **0.3125 moles** of KO2\mathrm{KO}_{2}.

STEP 9

Finally, we convert moles of KO2\mathrm{KO}_{2} to grams.
The molar mass of KO2\mathrm{KO}_{2} is 39.10+2⋅16.00=71.10 g/mol39.10 + 2 \cdot 16.00 = \mathbf{71.10 \text{ g/mol}}. grams of KO2=0.3125 moles⋅71.10 g/mol=22.21875 g\text{grams of } \mathrm{KO}_{2} = 0.3125 \text{ moles} \cdot 71.10 \text{ g/mol} = 22.21875 \text{ g}

STEP 10

(a) 0.300\mathbf{0.300} moles of O2\mathrm{O}_{2} are produced. (b) 22.2\mathbf{22.2} g of KO2\mathrm{KO}_{2} are needed.

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