Answer :
To determine the number of grams of oxygen gas ([tex]$\text{O}_2$[/tex]) required to react completely with 9.3 moles of aluminum ([tex]$\text{Al}$[/tex]), we can follow these detailed steps:
1. Understand the Balanced Chemical Equation:
[tex]\[ 4 \text{Al} + 3 \text{O}_2 \rightarrow 2 \text{Al}_2\text{O}_3 \][/tex]
The balanced equation tells us that 4 moles of aluminum react with 3 moles of oxygen gas.
2. Convert Given Information:
- We are given that we have 9.3 moles of aluminum ([tex]$\text{Al}$[/tex]).
- The molar mass of [tex]$\text{O}_2$[/tex] is 32.00 g/mol.
3. Use Stoichiometry to Find Moles of [tex]$\text{O}_2$[/tex] Needed:
- According to the balanced equation, [tex]$4$[/tex] moles of [tex]$\text{Al}$[/tex] react with [tex]$3$[/tex] moles of [tex]$\text{O}_2$[/tex].
- We need to determine how many moles of [tex]$\text{O}_2$[/tex] are required for [tex]$9.3$[/tex] moles of [tex]$\text{Al}$[/tex].
The ratio of moles of [tex]$\text{O}_2$[/tex] to moles of [tex]$\text{Al}$[/tex] according to the balanced equation is:
[tex]\[ \frac{3 \text{ moles of } \text{O}_2}{4 \text{ moles of } \text{Al}} \][/tex]
So, for [tex]$9.3$[/tex] moles of aluminum, the moles of [tex]$\text{O}_2$[/tex] needed are:
[tex]\[ \left(\frac{3}{4}\right) \times 9.3 \text{ moles of } \text{Al} = 6.975 \text{ moles of } \text{O}_2 \][/tex]
4. Convert Moles of [tex]$\text{O}_2$[/tex] to Grams:
- To find the mass of [tex]$\text{O}_2$[/tex] needed, we use the molar mass of [tex]$\text{O}_2$[/tex]:
[tex]\[ 6.975 \text{ moles of } \text{O}_2 \times 32.00 \text{ g/mol} = 223.2 \text{ grams of } \text{O}_2 \][/tex]
Final Answer:
To react completely with 9.3 moles of aluminum, we need 223.2 grams of oxygen gas ([tex]$\text{O}_2$[/tex]).
1. Understand the Balanced Chemical Equation:
[tex]\[ 4 \text{Al} + 3 \text{O}_2 \rightarrow 2 \text{Al}_2\text{O}_3 \][/tex]
The balanced equation tells us that 4 moles of aluminum react with 3 moles of oxygen gas.
2. Convert Given Information:
- We are given that we have 9.3 moles of aluminum ([tex]$\text{Al}$[/tex]).
- The molar mass of [tex]$\text{O}_2$[/tex] is 32.00 g/mol.
3. Use Stoichiometry to Find Moles of [tex]$\text{O}_2$[/tex] Needed:
- According to the balanced equation, [tex]$4$[/tex] moles of [tex]$\text{Al}$[/tex] react with [tex]$3$[/tex] moles of [tex]$\text{O}_2$[/tex].
- We need to determine how many moles of [tex]$\text{O}_2$[/tex] are required for [tex]$9.3$[/tex] moles of [tex]$\text{Al}$[/tex].
The ratio of moles of [tex]$\text{O}_2$[/tex] to moles of [tex]$\text{Al}$[/tex] according to the balanced equation is:
[tex]\[ \frac{3 \text{ moles of } \text{O}_2}{4 \text{ moles of } \text{Al}} \][/tex]
So, for [tex]$9.3$[/tex] moles of aluminum, the moles of [tex]$\text{O}_2$[/tex] needed are:
[tex]\[ \left(\frac{3}{4}\right) \times 9.3 \text{ moles of } \text{Al} = 6.975 \text{ moles of } \text{O}_2 \][/tex]
4. Convert Moles of [tex]$\text{O}_2$[/tex] to Grams:
- To find the mass of [tex]$\text{O}_2$[/tex] needed, we use the molar mass of [tex]$\text{O}_2$[/tex]:
[tex]\[ 6.975 \text{ moles of } \text{O}_2 \times 32.00 \text{ g/mol} = 223.2 \text{ grams of } \text{O}_2 \][/tex]
Final Answer:
To react completely with 9.3 moles of aluminum, we need 223.2 grams of oxygen gas ([tex]$\text{O}_2$[/tex]).
