Magnesium oxide can be made by heating magnesium metal in the presence of the oxygen. The balanced equation for the reaction is
2Mg(s)+O2(g)�¨2MgO(s)
When 10.2 g Mg is allowed to react with 10.6 g O2, 11.8 g MgO is collected.
Determine the theoretical yield for the reaction.
To determine the theoretical yield for the reaction, we need to calculate the amount of magnesium oxide that would be produced according to the balanced equation.
First, we need to convert the given masses of magnesium and oxygen to moles.
The molar mass of magnesium (Mg) is 24.31 g/mol, and the molar mass of oxygen (O2) is 32.00 g/mol.
Moles of magnesium = mass of magnesium / molar mass of magnesium
Moles of magnesium = 10.2 g / 24.31 g/mol ≈ 0.420 mol
Moles of oxygen = mass of oxygen / molar mass of oxygen
Moles of oxygen = 10.6 g / 32.00 g/mol ≈ 0.331 mol
From the balanced equation, we can see that the ratio of moles of magnesium to moles of magnesium oxide is 2:2, meaning that for every 2 moles of magnesium, 2 moles of magnesium oxide are produced.
Since we have 0.420 moles of magnesium, we can calculate the moles of magnesium oxide produced.
Moles of magnesium oxide = 0.420 mol × (2 mol MgO / 2 mol Mg) = 0.420 mol
Next, we need to convert the moles of magnesium oxide to grams using the molar mass of magnesium oxide (MgO), which is 40.31 g/mol.
Mass of magnesium oxide = moles of magnesium oxide × molar mass of magnesium oxide
Mass of magnesium oxide = 0.420 mol × 40.31 g/mol ≈ 16.93 g
Therefore, the theoretical yield for the reaction is approximately 16.93 g of magnesium oxide.
To determine the theoretical yield of the reaction, we need to calculate the maximum amount of product that can be formed based on the given amounts of reactants.
First, let's find the number of moles for each substance using their molar masses. The molar mass of magnesium (Mg) is 24.31 g/mol and the molar mass of oxygen (O2) is 32.00 g/mol.
For magnesium (Mg):
number of moles = mass / molar mass
number of moles of Mg = 10.2 g / 24.31 g/mol
number of moles of Mg = 0.420 moles
For oxygen (O2):
number of moles = mass / molar mass
number of moles of O2 = 10.6 g / 32.00 g/mol
number of moles of O2 = 0.331 moles
According to the balanced equation, the stoichiometric ratio between magnesium (Mg) and magnesium oxide (MgO) is 2:2, which means 2 moles of Mg reacts with 1 mole of O2 to produce 2 moles of MgO.
From the balanced equation, we can see that 2 moles of Mg react with 1 mole of O2 to produce 2 moles of MgO. Therefore, the limiting reactant is the one with a smaller number of moles.
In this case, oxygen (O2) is the limiting reactant because it has fewer moles than magnesium (Mg). The stoichiometric ratio tells us that for every 1 mole of O2, we will get 2 moles of MgO.
So, we first find the moles of MgO that can be formed from the limiting reactant:
moles of MgO = 0.331 moles O2 × (2 moles MgO / 1 mole O2)
moles of MgO = 0.662 moles
Now, we can find the maximum mass of MgO that can be formed:
mass of MgO = moles of MgO × molar mass of MgO
mass of MgO = 0.662 moles × 40.31 g/mol
mass of MgO = 26.623 g
Therefore, the theoretical yield of magnesium oxide (MgO) in this reaction is 26.623 grams.
This is a limiting reagent (LR) problem.
mols Mg = 10.2/24.3 = about 0.42 but you need a more accurate answer.
mols O2 = 11.8/32 = about 0.37 and
2 mols MgO/1 mol O2 = about 0.42/2 = about 0.21 and you have that much O2. Therefore, Mg must the LR.
mols Mg = 0.42
mols MgO produced = 0.42
g MgO = mols MgO x molar mass MgO = ?