Get comprehensive answers to your questions with the help of IDNLearn.com's community. Ask your questions and get detailed, reliable answers from our community of experienced experts.
Sagot :
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].
We value your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. IDNLearn.com is committed to your satisfaction. Thank you for visiting, and see you next time for more helpful answers.
