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To determine the maximum amount of magnesium oxide ([tex]\(\text{MgO}\)[/tex]) that can be produced during the reaction, follow these detailed steps:
Step 1: Determine the moles of reactants
First, calculate the number of moles of magnesium ([tex]\( \text{Mg} \)[/tex]) and oxygen ([tex]\( \text{O}_2 \)[/tex]).
For Magnesium ([tex]\( \text{Mg} \)[/tex]):
- The molar mass of [tex]\(\text{Mg}\)[/tex] is 24.305 grams per mole.
- Mass of [tex]\(\text{Mg}\)[/tex] is 1.8 grams.
[tex]\[ \text{Moles of Mg} = \frac{\text{mass of Mg}}{\text{molar mass of Mg}} = \frac{1.8 \text{ grams}}{24.305 \text{ grams/mol}} \approx 0.074 \text{ moles} \][/tex]
For Oxygen ([tex]\( \text{O}_2 \)[/tex]):
- The molar mass of [tex]\(\text{O}_2\)[/tex] is 32.00 grams per mole.
- Mass of [tex]\(\text{O}_2\)[/tex] is 6.0 grams.
[tex]\[ \text{Moles of O}_2 = \frac{\text{mass of O}_2}{\text{molar mass of O}_2} = \frac{6.0 \text{ grams}}{32.00 \text{ grams/mol}} = 0.1875 \text{ moles} \][/tex]
Step 2: Determine the limiting reactant
The stoichiometry of the reaction is as follows:
[tex]\[ 2 \text{Mg} + \text{O}_2 \rightarrow 2 \text{MgO} \][/tex]
From the stoichiometric ratio, 2 moles of [tex]\(\text{Mg}\)[/tex] react with 1 mole of [tex]\(\text{O}_2\)[/tex]. Therefore:
[tex]\[ \text{Moles of Mg} = 0.074 \][/tex]
[tex]\[ \text{Moles of O}_2 = 0.1875 \][/tex]
To find the limiting reactant, compare the actual mole ratio with the stoichiometric ratio:
[tex]\[ \text{Stoichiometry}:\, \frac{2 \text{ moles of Mg}}{1 \text{ mole of O}_2} \][/tex]
Based on our calculations, the moles of [tex]\(\text{Mg}\)[/tex] available is slightly less than twice the moles of [tex]\(\text{O}_2\)[/tex], so magnesium is the limiting reactant here.
Step 3: Calculate the maximum moles of [tex]\(\text{MgO}\)[/tex] that can be produced
From the stoichiometric relationship, 2 moles of [tex]\(\text{Mg}\)[/tex] produce 2 moles of [tex]\(\text{MgO}\)[/tex].
Therefore, the moles of [tex]\(\text{MgO}\)[/tex] produced are directly equal to the moles of [tex]\(\text{Mg}\)[/tex]:
[tex]\[ \text{Moles of MgO} = \text{Moles of Mg} = 0.074 \][/tex]
Step 4: Convert moles of [tex]\(\text{MgO}\)[/tex] to grams
Finally, calculate the mass of [tex]\(\text{MgO}\)[/tex] produced. The molar mass of [tex]\(\text{MgO}\)[/tex] is 40.305 grams per mole.
[tex]\[ \text{Mass of MgO} = \text{moles of MgO} \times \text{molar mass of MgO} = 0.074 \times 40.305 \approx 2.985 \text{ grams} \][/tex]
So, the maximum amount of magnesium oxide that can be produced during the reaction is approximately 2.985 grams.
Given the choices:
- 2.2 grams
- 2.9 grams
- 3.4 grams
- 3.9 grams
The closest value is 2.9 grams.
Thus, the maximum amount of magnesium oxide that can be produced is [tex]\(\boxed{2.9 \text{ grams}}\)[/tex].
Step 1: Determine the moles of reactants
First, calculate the number of moles of magnesium ([tex]\( \text{Mg} \)[/tex]) and oxygen ([tex]\( \text{O}_2 \)[/tex]).
For Magnesium ([tex]\( \text{Mg} \)[/tex]):
- The molar mass of [tex]\(\text{Mg}\)[/tex] is 24.305 grams per mole.
- Mass of [tex]\(\text{Mg}\)[/tex] is 1.8 grams.
[tex]\[ \text{Moles of Mg} = \frac{\text{mass of Mg}}{\text{molar mass of Mg}} = \frac{1.8 \text{ grams}}{24.305 \text{ grams/mol}} \approx 0.074 \text{ moles} \][/tex]
For Oxygen ([tex]\( \text{O}_2 \)[/tex]):
- The molar mass of [tex]\(\text{O}_2\)[/tex] is 32.00 grams per mole.
- Mass of [tex]\(\text{O}_2\)[/tex] is 6.0 grams.
[tex]\[ \text{Moles of O}_2 = \frac{\text{mass of O}_2}{\text{molar mass of O}_2} = \frac{6.0 \text{ grams}}{32.00 \text{ grams/mol}} = 0.1875 \text{ moles} \][/tex]
Step 2: Determine the limiting reactant
The stoichiometry of the reaction is as follows:
[tex]\[ 2 \text{Mg} + \text{O}_2 \rightarrow 2 \text{MgO} \][/tex]
From the stoichiometric ratio, 2 moles of [tex]\(\text{Mg}\)[/tex] react with 1 mole of [tex]\(\text{O}_2\)[/tex]. Therefore:
[tex]\[ \text{Moles of Mg} = 0.074 \][/tex]
[tex]\[ \text{Moles of O}_2 = 0.1875 \][/tex]
To find the limiting reactant, compare the actual mole ratio with the stoichiometric ratio:
[tex]\[ \text{Stoichiometry}:\, \frac{2 \text{ moles of Mg}}{1 \text{ mole of O}_2} \][/tex]
Based on our calculations, the moles of [tex]\(\text{Mg}\)[/tex] available is slightly less than twice the moles of [tex]\(\text{O}_2\)[/tex], so magnesium is the limiting reactant here.
Step 3: Calculate the maximum moles of [tex]\(\text{MgO}\)[/tex] that can be produced
From the stoichiometric relationship, 2 moles of [tex]\(\text{Mg}\)[/tex] produce 2 moles of [tex]\(\text{MgO}\)[/tex].
Therefore, the moles of [tex]\(\text{MgO}\)[/tex] produced are directly equal to the moles of [tex]\(\text{Mg}\)[/tex]:
[tex]\[ \text{Moles of MgO} = \text{Moles of Mg} = 0.074 \][/tex]
Step 4: Convert moles of [tex]\(\text{MgO}\)[/tex] to grams
Finally, calculate the mass of [tex]\(\text{MgO}\)[/tex] produced. The molar mass of [tex]\(\text{MgO}\)[/tex] is 40.305 grams per mole.
[tex]\[ \text{Mass of MgO} = \text{moles of MgO} \times \text{molar mass of MgO} = 0.074 \times 40.305 \approx 2.985 \text{ grams} \][/tex]
So, the maximum amount of magnesium oxide that can be produced during the reaction is approximately 2.985 grams.
Given the choices:
- 2.2 grams
- 2.9 grams
- 3.4 grams
- 3.9 grams
The closest value is 2.9 grams.
Thus, the maximum amount of magnesium oxide that can be produced is [tex]\(\boxed{2.9 \text{ grams}}\)[/tex].
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