Answer :
Sure! Let's go through the step-by-step solution to determine how many grams of [tex]\( \text{Fe}_2\text{O}_3 \)[/tex] can be produced from 38.6 moles of Fe based on the chemical reaction provided:
Chemical Reaction:
[tex]\[ 4 \text{Fe} + 3 \text{O}_2 \rightarrow 2 \text{Fe}_2\text{O}_3 \][/tex]
Step 1: Understand the Stoichiometry of the Reaction
- The balanced equation shows that 4 moles of iron ([tex]\(\text{Fe}\)[/tex]) produce 2 moles of iron(III) oxide ([tex]\(\text{Fe}_2\text{O}_3\)[/tex]).
Step 2: Calculate the Moles of [tex]\( \text{Fe}_2\text{O}_3 \)[/tex] Produced
- Given that you have 38.6 moles of [tex]\(\text{Fe}\)[/tex], and knowing that the proportion is 4 moles of [tex]\(\text{Fe}\)[/tex] to 2 moles of [tex]\(\text{Fe}_2\text{O}_3\)[/tex], you can set up the ratio:
[tex]\[
\text{Moles of } \text{Fe}_2\text{O}_3 = \left(\frac{2}{4}\right) \times \text{Moles of Fe} = \frac{1}{2} \times 38.6 = 19.3 \text{ moles}
\][/tex]
Step 3: Calculate the Mass of [tex]\( \text{Fe}_2\text{O}_3 \)[/tex] Produced
- The molar mass of [tex]\(\text{Fe}_2\text{O}_3\)[/tex] is given as 159.70 g/mol.
- To find out how many grams of [tex]\(\text{Fe}_2\text{O}_3\)[/tex] are produced, multiply the moles of [tex]\(\text{Fe}_2\text{O}_3\)[/tex] by its molar mass:
[tex]\[
\text{Grams of } \text{Fe}_2\text{O}_3 = 19.3 \text{ moles} \times 159.70 \text{ g/mol} = 3082.21 \text{ grams}
\][/tex]
Therefore, 38.6 moles of Fe can produce approximately 3082.21 grams of [tex]\( \text{Fe}_2\text{O}_3 \)[/tex].
Chemical Reaction:
[tex]\[ 4 \text{Fe} + 3 \text{O}_2 \rightarrow 2 \text{Fe}_2\text{O}_3 \][/tex]
Step 1: Understand the Stoichiometry of the Reaction
- The balanced equation shows that 4 moles of iron ([tex]\(\text{Fe}\)[/tex]) produce 2 moles of iron(III) oxide ([tex]\(\text{Fe}_2\text{O}_3\)[/tex]).
Step 2: Calculate the Moles of [tex]\( \text{Fe}_2\text{O}_3 \)[/tex] Produced
- Given that you have 38.6 moles of [tex]\(\text{Fe}\)[/tex], and knowing that the proportion is 4 moles of [tex]\(\text{Fe}\)[/tex] to 2 moles of [tex]\(\text{Fe}_2\text{O}_3\)[/tex], you can set up the ratio:
[tex]\[
\text{Moles of } \text{Fe}_2\text{O}_3 = \left(\frac{2}{4}\right) \times \text{Moles of Fe} = \frac{1}{2} \times 38.6 = 19.3 \text{ moles}
\][/tex]
Step 3: Calculate the Mass of [tex]\( \text{Fe}_2\text{O}_3 \)[/tex] Produced
- The molar mass of [tex]\(\text{Fe}_2\text{O}_3\)[/tex] is given as 159.70 g/mol.
- To find out how many grams of [tex]\(\text{Fe}_2\text{O}_3\)[/tex] are produced, multiply the moles of [tex]\(\text{Fe}_2\text{O}_3\)[/tex] by its molar mass:
[tex]\[
\text{Grams of } \text{Fe}_2\text{O}_3 = 19.3 \text{ moles} \times 159.70 \text{ g/mol} = 3082.21 \text{ grams}
\][/tex]
Therefore, 38.6 moles of Fe can produce approximately 3082.21 grams of [tex]\( \text{Fe}_2\text{O}_3 \)[/tex].
