To solve the problem of finding out how many grams of oxygen gas (O₂) are required to completely react with 9.30 moles of aluminum (Al) according to the balanced chemical equation:
[tex]\[ 4 \text{Al} + 3 \text{O}_2 \rightarrow 2 \text{Al}_2\text{O}_3 \][/tex]
we will follow these steps:
1. Determine the UNKNOWN:
- The unknown in the problem is the grams of [tex]\( \text{O}_2 \)[/tex] needed.
2. Identify the given values:
- Moles of Al given: 9.30 moles
3. Use the stoichiometric coefficients from the balanced equation to find the moles of [tex]\( \text{O}_2 \)[/tex] required:
- According to the balanced equation, 4 moles of Al react with 3 moles of [tex]\( \text{O}_2 \)[/tex].
- The ratio can be expressed as:
[tex]\[
\frac{3 \, \text{moles} \, \text{O}_2}{4 \, \text{moles} \, \text{Al}}
\][/tex]
4. Calculate the moles of [tex]\( \text{O}_2 \)[/tex] needed:
- Multiply the given moles of Al by the stoichiometric ratio to get the moles of [tex]\( \text{O}_2 \)[/tex]:
[tex]\[
\text{Moles of } \text{O}_2 = 9.30 \, \text{moles} \, \text{Al} \times \frac{3 \, \text{moles} \, \text{O}_2}{4 \, \text{moles} \, \text{Al}} = 6.975 \, \text{moles} \, \text{O}_2
\][/tex]
5. Convert moles of [tex]\( \text{O}_2 \)[/tex] to grams:
- Use the molar mass of [tex]\( \text{O}_2 \)[/tex] to convert to grams. The molar mass of [tex]\( \text{O}_2 \)[/tex] is 32 g/mol.
- Multiplying the moles of [tex]\( \text{O}_2 \)[/tex] by its molar mass:
[tex]\[
\text{Grams of } \text{O}_2 = 6.975 \, \text{moles} \, \text{O}_2 \times 32 \, \text{g/mol} = 223.2 \, \text{grams}
\][/tex]
Hence, the amount of oxygen gas [tex]\( \text{O}_2 \)[/tex] needed to completely react with 9.30 moles of aluminum is 6.975 moles of [tex]\( \text{O}_2 \)[/tex] and 223.2 grams of [tex]\( \text{O}_2 \)[/tex].
