For the reaction 2Zn plus O2>> 2ZnO, how many grams of zinc oxide can be produced from 100.g each of zinc and oxygen?
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To calculate the amount of zinc oxide produced, we need to determine the limiting reactant in the reaction. The limiting reactant is the reactant that is completely consumed and limits the amount of product that can be formed.
Let's start by calculating the number of moles of zinc and oxygen:
Molar mass of zinc (Zn) = 65.38 g/mol
Molar mass of oxygen (O2) = 32.00 g/mol
Number of moles of zinc = mass of zinc / molar mass of zinc
Number of moles of zinc = 100 g / 65.38 g/mol ≈ 1.529 mol
Number of moles of oxygen = mass of oxygen / molar mass of oxygen
Number of moles of oxygen = 100 g / 32.00 g/mol ≈ 3.125 mol
According to the balanced equation, the stoichiometric ratio between zinc and zinc oxide is 2:2. Therefore, both reactants have a 1:1 mole ratio with zinc oxide.
As we have 1.529 moles of zinc and 3.125 moles of oxygen, we can see that the zinc is the limiting reactant since we have less moles of zinc compared to oxygen.
Next, we need to calculate the moles of zinc oxide produced using the limiting reactant:
Number of moles of zinc oxide = limiting reactant moles * (1 mole of ZnO / 2 moles of Zn)
Number of moles of zinc oxide = 1.529 mol * (1 mol ZnO / 2 mol Zn) ≈ 0.7645 mol
Now, we can determine the mass of zinc oxide produced:
Mass of zinc oxide = moles of zinc oxide * molar mass of zinc oxide
Mass of zinc oxide = 0.7645 mol * 81.38 g/mol ≈ 62.24 g
Therefore, approximately 62.24 grams of zinc oxide can be produced from 100 g each of zinc and oxygen.
To determine the amount of zinc oxide that can be produced from the given amounts of zinc and oxygen, we need to follow these steps:
1. Write down the balanced chemical equation for the reaction:
2Zn + O2 → 2ZnO
2. Calculate the molar masses of the reactants and the product:
- Molar mass of Zn (zinc) = 65.38 g/mol
- Molar mass of O2 (oxygen) = 32.00 g/mol
- Molar mass of ZnO (zinc oxide) = 81.38 g/mol (65.38 g/mol for Zn + 16.00 g/mol for O)
3. Convert the given mass of zinc (100. g) to moles:
Number of moles of Zn = mass of Zn / molar mass of Zn
Number of moles of Zn = 100. g / 65.38 g/mol
Number of moles of Zn = 1.529 mol
4. Convert the given mass of oxygen (100. g) to moles:
Number of moles of O2 = mass of O2 / molar mass of O2
Number of moles of O2 = 100. g / 32.00 g/mol
Number of moles of O2 = 3.125 mol
5. Determine the limiting reactant:
In this case, we have a 2:1 molar ratio between Zn and O2. So, for every 2 moles of Zn, we need 1 mole of O2. However, we have 1.529 moles of Zn and 3.125 moles of O2. Since we have an excess of oxygen, Zn is the limiting reactant.
6. Calculate the theoretical yield of ZnO using the limiting reactant:
Using the balanced equation, we see that 2 moles of Zn produce 2 moles of ZnO.
So, 1.529 moles of Zn will produce (1.529 mol * 2 mol ZnO) / 2 mol Zn = 1.529 mol of ZnO
7. Convert the moles of ZnO to grams:
Mass of ZnO = number of moles of ZnO * molar mass of ZnO
Mass of ZnO = 1.529 mol * 81.38 g/mol
Mass of ZnO = 124.53 g
Therefore, from 100 g of zinc and oxygen, the maximum amount of zinc oxide that can be produced is 124.53 grams.