Question

part 2: write out and balance: iron and oxygen react to form iron (iii) oxide. then answer questions3. determine how many grams of $ce{fe_{2}o_{3}}$ you will make if you with 2.0 mol fe and 6.0 mol $ce{o_{2}}$4. determine how many molecules of $ce{fe_{2}o_{3}}$ you will make if you with 37.1 g fe and $2.45 \times 10^{24}$ molecules of $ce{o_{2}}$

Explanation:

Step1: Balance the reaction equation

Unbalanced: $\text{Fe} + \text{O}_2
ightarrow \text{Fe}_2\text{O}_3$
Balanced: $4\text{Fe} + 3\text{O}_2
ightarrow 2\text{Fe}_2\text{O}_3$

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For Question 3:

Step2: Identify limiting reactant

From balanced equation, mole ratio $\text{Fe}:\text{Fe}_2\text{O}_3 = 4:2 = 2:1$
Moles of $\text{Fe}_2\text{O}_3$ from $\text{Fe}$: $\frac{2.0\ \text{mol}}{2} = 1.0\ \text{mol}$

Mole ratio $\text{O}_2:\text{Fe}_2\text{O}_3 = 3:2$
Moles of $\text{Fe}_2\text{O}_3$ from $\text{O}_2$: $6.0\ \text{mol} \times \frac{2}{3} = 4.0\ \text{mol}$
Fe is the limiting reactant, so use 1.0 mol $\text{Fe}_2\text{O}_3$.

Step3: Calculate molar mass of $\text{Fe}_2\text{O}_3$

Molar mass: $2\times55.85 + 3\times16.00 = 159.7\ \text{g/mol}$

Step4: Compute mass of $\text{Fe}_2\text{O}_3$

Mass = moles $\times$ molar mass
$\text{Mass} = 1.0\ \text{mol} \times 159.7\ \text{g/mol} = 159.7\ \text{g}$

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For Question 4:

Step5: Convert Fe mass to moles

Moles of $\text{Fe} = \frac{37.1\ \text{g}}{55.85\ \text{g/mol}} \approx 0.664\ \text{mol}$

Step6: Convert $\text{O}_2$ molecules to moles

Moles of $\text{O}_2 = \frac{2.45\times10^{24}}{6.022\times10^{23}\ \text{mol}^{-1}} \approx 4.07\ \text{mol}$

Step7: Find limiting reactant

Moles of $\text{Fe}_2\text{O}_3$ from $\text{Fe}$: $\frac{0.664\ \text{mol}}{2} = 0.332\ \text{mol}$
Moles of $\text{Fe}_2\text{O}_3$ from $\text{O}_2$: $4.07\ \text{mol} \times \frac{2}{3} \approx 2.71\ \text{mol}$
Fe is the limiting reactant, so use 0.332 mol $\text{Fe}_2\text{O}_3$.

Step8: Convert moles to molecules

Molecules = moles $\times N_A$
$\text{Molecules} = 0.332\ \text{mol} \times 6.022\times10^{23}\ \text{mol}^{-1} \approx 2.00\times10^{23}$

Answer:

1. $160\ \text{g}$ (rounded to 2 significant figures)
2. $2.00\times10^{23}$ molecules