Answer :
Certainly! Let's go through the problem step by step.
Step 1: Understand the given data and balanced chemical equation
The balanced chemical equation is:
[tex]\[ 4 \text{Fe (s)} + 3 \text{O}_2 \text{(g)} \rightarrow 2 \text{Fe}_2\text{O}_3 \text{(s)} \][/tex]
Step 2: Determine the molar masses
From the periodic table:
- The molar mass of [tex]\( \text{O}_2 \)[/tex] is approximately [tex]\( 32.00 \, \text{g/mol} \)[/tex].
- The molar mass of [tex]\( \text{Fe}_2\text{O}_3 \)[/tex] is approximately [tex]\( 159.69 \, \text{g/mol} \)[/tex].
Step 3: Calculate the number of moles of [tex]\( \text{O}_2 \)[/tex]
Using the formula to convert grams to moles:
[tex]\[ \text{moles of } \text{O}_2 = \frac{\text{mass of } \text{O}_2}{\text{molar mass of } \text{O}_2} \][/tex]
[tex]\[ \text{moles of } \text{O}_2 = \frac{26.7 \, \text{g}}{32.00 \, \text{g/mol}} = 0.834375 \, \text{moles} \][/tex]
Step 4: Use stoichiometry to find the moles of [tex]\( \text{Fe}_2\text{O}_3 \)[/tex]
From the balanced chemical equation, we know that 3 moles of [tex]\( \text{O}_2 \)[/tex] produce 2 moles of [tex]\( \text{Fe}_2\text{O}_3 \)[/tex]. So, we can set up the following ratio:
[tex]\[ \frac{2 \, \text{moles of } \text{Fe}_2\text{O}_3}{3 \, \text{moles of } \text{O}_2} \][/tex]
Now, we use this ratio to find out how many moles of [tex]\( \text{Fe}_2\text{O}_3 \)[/tex] are formed from the 0.834375 moles of [tex]\( \text{O}_2 \)[/tex]:
[tex]\[ \text{moles of } \text{Fe}_2\text{O}_3 = 0.834375 \, \text{moles of } \text{O}_2 \times \frac{2 \, \text{moles of } \text{Fe}_2\text{O}_3}{3 \, \text{moles of } \text{O}_2} \][/tex]
[tex]\[ \text{moles of } \text{Fe}_2\text{O}_3 = 0.55625 \, \text{moles} \][/tex]
Step 5: Convert moles of [tex]\( \text{Fe}_2\text{O}_3 \)[/tex] to mass
Using the molar mass of [tex]\( \text{Fe}_2\text{O}_3 \)[/tex]:
[tex]\[ \text{mass of } \text{Fe}_2\text{O}_3 = \text{moles of } \text{Fe}_2\text{O}_3 \times \text{molar mass of } \text{Fe}_2\text{O}_3 \][/tex]
[tex]\[ \text{mass of } \text{Fe}_2\text{O}_3 = 0.55625 \, \text{moles} \times 159.69 \, \text{g/mol} \][/tex]
[tex]\[ \text{mass of } \text{Fe}_2\text{O}_3 = 88.8275625 \, \text{g} \][/tex]
Final Answer: The mass of [tex]\( \text{Fe}_2\text{O}_3 \)[/tex] that can form from 26.7 grams of [tex]\( \text{O}_2 \)[/tex] is approximately 88.83 grams.
Step 1: Understand the given data and balanced chemical equation
The balanced chemical equation is:
[tex]\[ 4 \text{Fe (s)} + 3 \text{O}_2 \text{(g)} \rightarrow 2 \text{Fe}_2\text{O}_3 \text{(s)} \][/tex]
Step 2: Determine the molar masses
From the periodic table:
- The molar mass of [tex]\( \text{O}_2 \)[/tex] is approximately [tex]\( 32.00 \, \text{g/mol} \)[/tex].
- The molar mass of [tex]\( \text{Fe}_2\text{O}_3 \)[/tex] is approximately [tex]\( 159.69 \, \text{g/mol} \)[/tex].
Step 3: Calculate the number of moles of [tex]\( \text{O}_2 \)[/tex]
Using the formula to convert grams to moles:
[tex]\[ \text{moles of } \text{O}_2 = \frac{\text{mass of } \text{O}_2}{\text{molar mass of } \text{O}_2} \][/tex]
[tex]\[ \text{moles of } \text{O}_2 = \frac{26.7 \, \text{g}}{32.00 \, \text{g/mol}} = 0.834375 \, \text{moles} \][/tex]
Step 4: Use stoichiometry to find the moles of [tex]\( \text{Fe}_2\text{O}_3 \)[/tex]
From the balanced chemical equation, we know that 3 moles of [tex]\( \text{O}_2 \)[/tex] produce 2 moles of [tex]\( \text{Fe}_2\text{O}_3 \)[/tex]. So, we can set up the following ratio:
[tex]\[ \frac{2 \, \text{moles of } \text{Fe}_2\text{O}_3}{3 \, \text{moles of } \text{O}_2} \][/tex]
Now, we use this ratio to find out how many moles of [tex]\( \text{Fe}_2\text{O}_3 \)[/tex] are formed from the 0.834375 moles of [tex]\( \text{O}_2 \)[/tex]:
[tex]\[ \text{moles of } \text{Fe}_2\text{O}_3 = 0.834375 \, \text{moles of } \text{O}_2 \times \frac{2 \, \text{moles of } \text{Fe}_2\text{O}_3}{3 \, \text{moles of } \text{O}_2} \][/tex]
[tex]\[ \text{moles of } \text{Fe}_2\text{O}_3 = 0.55625 \, \text{moles} \][/tex]
Step 5: Convert moles of [tex]\( \text{Fe}_2\text{O}_3 \)[/tex] to mass
Using the molar mass of [tex]\( \text{Fe}_2\text{O}_3 \)[/tex]:
[tex]\[ \text{mass of } \text{Fe}_2\text{O}_3 = \text{moles of } \text{Fe}_2\text{O}_3 \times \text{molar mass of } \text{Fe}_2\text{O}_3 \][/tex]
[tex]\[ \text{mass of } \text{Fe}_2\text{O}_3 = 0.55625 \, \text{moles} \times 159.69 \, \text{g/mol} \][/tex]
[tex]\[ \text{mass of } \text{Fe}_2\text{O}_3 = 88.8275625 \, \text{g} \][/tex]
Final Answer: The mass of [tex]\( \text{Fe}_2\text{O}_3 \)[/tex] that can form from 26.7 grams of [tex]\( \text{O}_2 \)[/tex] is approximately 88.83 grams.
