Stoichiometry practice problems involving mole-to-mole conversions for various chemical reactions.

To solve these stoichiometry problems, we need to use the balanced chemical equations provided for each section. The coefficients in the balanced equations tell us the mole ratios between the reactants and products. We will use these ratios to convert from one substance to another.

Section 1: $2 \text{ KClO}_3 \rightarrow 2 \text{ KCl} + 3 \text{ O}_2$

* Problem 1: How many moles of $\text{O}_2$ will be formed from $1.65$ moles of $\text{KClO}_3$?
* From the equation, $2$ moles of $\text{KClO}_3$ produce $3$ moles of $\text{O}_2$.
* Ratio: $\frac{3 \text{ mol O}_2}{2 \text{ mol KClO}_3}$
* Calculation: $1.65 \text{ mol KClO}_3 \times \frac{3}{2} = 1.65 \times 1.5 = 2.475 \text{ mol O}_2$

* Problem 2: How many moles of $\text{KClO}_3$ are needed to make $3.50$ moles of $\text{KCl}$?
* From the equation, $2$ moles of $\text{KClO}_3$ produce $2$ moles of $\text{KCl}$ (a 1:1 ratio).
* Ratio: $\frac{2 \text{ mol KClO}_3}{2 \text{ mol KCl}}$
* Calculation: $3.50 \text{ mol KCl} \times \frac{2}{2} = 3.50 \text{ mol KClO}_3$

* Problem 3: How many moles of $\text{KCl}$ will be formed from $2.73$ moles of $\text{KClO}_3$?
* From the equation, $2$ moles of $\text{KClO}_3$ produce $2$ moles of $\text{KCl}$ (a 1:1 ratio).
* Ratio: $\frac{2 \text{ mol KCl}}{2 \text{ mol KClO}_3}$
* Calculation: $2.73 \text{ mol KClO}_3 \times \frac{2}{2} = 2.73 \text{ mol KCl}$

Section 2: $4 \text{ Fe} + 3 \text{ O}_2 \rightarrow 2 \text{ Fe}_2\text{O}_3$

* Problem 4: How many moles of $\text{Fe}_2\text{O}_3$ are produced when $0.275$ moles of $\text{Fe}$ is reacted?
* From the equation, $4$ moles of $\text{Fe}$ produce $2$ moles of $\text{Fe}_2\text{O}_3$.
* Ratio: $\frac{2 \text{ mol Fe}_2\text{O}_3}{4 \text{ mol Fe}} = \frac{1}{2}$
* Calculation: $0.275 \text{ mol Fe} \times \frac{2}{4} = 0.275 \times 0.5 = 0.1375 \text{ mol Fe}_2\text{O}_3$

* Problem 5: How many moles of $\text{Fe}_2\text{O}_3$ are produced when $31.0$ moles of $\text{O}_2$ is reacted?
* From the equation, $3$ moles of $\text{O}_2$ produce $2$ moles of $\text{Fe}_2\text{O}_3$.
* Ratio: $\frac{2 \text{ mol Fe}_2\text{O}_3}{3 \text{ mol O}_2}$
* Calculation: $31.0 \text{ mol O}_2 \times \frac{2}{3} = \frac{62}{3} \approx 20.67 \text{ mol Fe}_2\text{O}_3$

* Problem 6: How many moles of $\text{O}_2$ are needed to react with $8.9$ moles of $\text{Fe}$?
* From the equation, $4$ moles of $\text{Fe}$ react with $3$ moles of $\text{O}_2$.
* Ratio: $\frac{3 \text{ mol O}_2}{4 \text{ mol Fe}}$
* Calculation: $8.9 \text{ mol Fe} \times \frac{3}{4} = 8.9 \times 0.75 = 6.675 \text{ mol O}_2$

Section 3: $2 \text{ H}_2\text{O} \rightarrow 2 \text{ H}_2 + \text{O}_2$

* Problem 7: How many moles of $\text{O}_2$ are produced when $1.26$ moles of $\text{H}_2\text{O}$ is reacted?
* From the equation, $2$ moles of $\text{H}_2\text{O}$ produce $1$ mole of $\text{O}_2$.
* Ratio: $\frac{1 \text{ mol O}_2}{2 \text{ mol H}_2\text{O}}$
* Calculation: $1.26 \text{ mol H}_2\text{O} \times \frac{1}{2} = 0.63 \text{ mol O}_2$

* Problem 8: How many moles of $\text{H}_2\text{O}$ are needed to produce $55.7$ moles of $\text{H}_2$?
* From the equation, $2$ moles of $\text{H}_2\text{O}$ produce $2$ moles of $\text{H}_2$ (a 1:1 ratio).
* Ratio: $\frac{2 \text{ mol H}_2\text{O}}{2 \text{ mol H}_2}$
* Calculation: $55.7 \text{ mol H}_2 \times \frac{2}{2} = 55.7 \text{ mol H}_2\text{O}$

* Problem 9: If enough $\text{H}_2\text{O}$ is reacted to produce $3.40$ moles of $\text{H}_2$, then how may moles of $\text{O}_2$ must have been made?
* Look at the product side of the equation: $2 \text{ H}_2 + 1 \text{ O}_2$.
* For every $2$ moles of $\text{H}_2$ produced, $1$ mole of $\text{O}_2$ is also produced.
* Ratio: $\frac{1 \text{ mol O}_2}{2 \text{ mol H}_2}$
* Calculation: $3.40 \text{ mol H}_2 \times \frac{1}{2} = 1.70 \text{ mol O}_2$

Final Answer:
1. 2.475 mol O₂
2. 3.50 mol KClO₃
3. 2.73 mol KCl
4. 0.1375 mol Fe₂O₃
5. 20.67 mol Fe₂O₃
6. 6.675 mol O₂
7. 0.63 mol O₂
8. 55.7 mol H₂O
9. 1.70 mol O₂