Answer :
To determine how many moles of zinc oxide ([tex]\( ZnO \)[/tex]) can be produced from 100 grams each of zinc ([tex]\( Zn \)[/tex]) and oxygen ([tex]\( O_2 \)[/tex]), let's follow these steps:
1. Finding the amounts in moles:
- The molar mass of zinc ([tex]\( Zn \)[/tex]) is 65.38 g/mol.
- The molar mass of oxygen ([tex]\( O_2 \)[/tex]) is 32.00 g/mol.
First, we convert the masses of zinc and oxygen to moles:
- Moles of [tex]\( Zn \)[/tex]:
[tex]\[ \text{Moles of } Zn = \frac{100.0 \text{ g}}{65.38 \text{ g/mol}} \approx 1.53 \text{ moles} \][/tex]
- Moles of [tex]\( O_2 \)[/tex]:
[tex]\[ \text{Moles of } O_2 = \frac{100.0 \text{ g}}{32.00 \text{ g/mol}} = 3.125 \text{ moles} \][/tex]
2. Identifying the limiting reagent:
The balanced chemical equation for the reaction is:
[tex]\[ 2Zn + O_2 \rightarrow 2ZnO \][/tex]
According to the reaction equation, two moles of [tex]\( Zn \)[/tex] react with one mole of [tex]\( O_2 \)[/tex] to produce two moles of [tex]\( ZnO \)[/tex].
To determine the limiting reagent, we need to see which reactant will run out first based on the stoichiometric ratios:
- Moles of [tex]\( Zn \)[/tex] required for the available [tex]\( O_2 \)[/tex]:
[tex]\[ 2 \text{ moles of } Zn \text{ are needed for every 1 mole of } O_2 \][/tex]
[tex]\[ \text{Moles of } Zn \text{ needed} = 3.125 \text{ moles of } O_2 \times 2 = 6.25 \text{ moles of } Zn \][/tex]
We only have approximately 1.53 moles of [tex]\( Zn \)[/tex], which is less than 6.25 moles. Therefore, [tex]\( Zn \)[/tex] is the limiting reagent.
3. Calculating the number of moles of [tex]\( ZnO \)[/tex] produced:
Since [tex]\( Zn \)[/tex] is the limiting reagent, let's use its amount to determine the moles of [tex]\( ZnO \)[/tex] produced. According to the reaction equation:
- 2 moles of [tex]\( Zn \)[/tex] produce 2 moles of [tex]\( ZnO \)[/tex].
Therefore, the moles of [tex]\( ZnO \)[/tex] produced will be equal to the moles of [tex]\( Zn \)[/tex] we have:
[tex]\[ \text{Moles of } ZnO \text{ produced} = \text{Moles of } Zn = 1.53 \text{ moles} \][/tex]
So, the detailed solution gives us the following:
- Moles of [tex]\( Zn \)[/tex]: [tex]\( \approx 1.53 \)[/tex]
- Moles of [tex]\( O_2 \)[/tex]: [tex]\( 3.125 \)[/tex]
- Moles of [tex]\( ZnO \)[/tex] produced: [tex]\( \approx 1.53 \)[/tex]
Thus, approximately 1.53 moles of zinc oxide can be produced from 100 grams each of zinc and oxygen.
1. Finding the amounts in moles:
- The molar mass of zinc ([tex]\( Zn \)[/tex]) is 65.38 g/mol.
- The molar mass of oxygen ([tex]\( O_2 \)[/tex]) is 32.00 g/mol.
First, we convert the masses of zinc and oxygen to moles:
- Moles of [tex]\( Zn \)[/tex]:
[tex]\[ \text{Moles of } Zn = \frac{100.0 \text{ g}}{65.38 \text{ g/mol}} \approx 1.53 \text{ moles} \][/tex]
- Moles of [tex]\( O_2 \)[/tex]:
[tex]\[ \text{Moles of } O_2 = \frac{100.0 \text{ g}}{32.00 \text{ g/mol}} = 3.125 \text{ moles} \][/tex]
2. Identifying the limiting reagent:
The balanced chemical equation for the reaction is:
[tex]\[ 2Zn + O_2 \rightarrow 2ZnO \][/tex]
According to the reaction equation, two moles of [tex]\( Zn \)[/tex] react with one mole of [tex]\( O_2 \)[/tex] to produce two moles of [tex]\( ZnO \)[/tex].
To determine the limiting reagent, we need to see which reactant will run out first based on the stoichiometric ratios:
- Moles of [tex]\( Zn \)[/tex] required for the available [tex]\( O_2 \)[/tex]:
[tex]\[ 2 \text{ moles of } Zn \text{ are needed for every 1 mole of } O_2 \][/tex]
[tex]\[ \text{Moles of } Zn \text{ needed} = 3.125 \text{ moles of } O_2 \times 2 = 6.25 \text{ moles of } Zn \][/tex]
We only have approximately 1.53 moles of [tex]\( Zn \)[/tex], which is less than 6.25 moles. Therefore, [tex]\( Zn \)[/tex] is the limiting reagent.
3. Calculating the number of moles of [tex]\( ZnO \)[/tex] produced:
Since [tex]\( Zn \)[/tex] is the limiting reagent, let's use its amount to determine the moles of [tex]\( ZnO \)[/tex] produced. According to the reaction equation:
- 2 moles of [tex]\( Zn \)[/tex] produce 2 moles of [tex]\( ZnO \)[/tex].
Therefore, the moles of [tex]\( ZnO \)[/tex] produced will be equal to the moles of [tex]\( Zn \)[/tex] we have:
[tex]\[ \text{Moles of } ZnO \text{ produced} = \text{Moles of } Zn = 1.53 \text{ moles} \][/tex]
So, the detailed solution gives us the following:
- Moles of [tex]\( Zn \)[/tex]: [tex]\( \approx 1.53 \)[/tex]
- Moles of [tex]\( O_2 \)[/tex]: [tex]\( 3.125 \)[/tex]
- Moles of [tex]\( ZnO \)[/tex] produced: [tex]\( \approx 1.53 \)[/tex]
Thus, approximately 1.53 moles of zinc oxide can be produced from 100 grams each of zinc and oxygen.
