Answer:
163.2g
Explanation:
First let us generate a balanced equation for the reaction. This is shown below:
4Al + 3O2 —> 2Al2O3
From the question given, were were told that 3.2moles of aluminium was exposed to 2.7moles of oxygen. Judging by this, oxygen is excess.
From the equation,
4moles of Al produced 2moles of Al2O3.
Therefore, 3.2moles of Al will produce = (3.2x2)/4 = 1.6mol of Al2O3.
Now, let us covert 1.6mol of Al2O3 to obtain the theoretical yield. This is illustrated below:
Mole of Al2O3 = 1.6mole
Molar Mass of Al2O3 = (27x2) + (16x3) = 54 + 48 =102g/mol
Mass of Al2O3 =?
Number of mole = Mass /Molar Mass
Mass = number of mole x molar Mass
Mass of Al2O3 = 1.6 x 102 = 163.2g
Therefore the theoretical of Al2O3 is 163.2g
Answer:
Explanation:
You don't give the reaction, but we can get by just by balancing atoms of Na.
We know we will need the partially balanced equation with masses, moles, and molar masses, so let’s gather all the information in one place.
M_r: 142.04
2NaOH + … ⟶ Na₂SO₄ + …
n/mol: 0.75
1. Use the molar ratio of Na₂SO₄ to NaOH to calculate the moles of NaF.
Moles of Na₂SO₄ = 0.75 mol NaOH × (1 mol Na₂SO₄/2 mol NaOH
= 0.375 mol Na₂SO₄
2. Use the molar mass of Na₂SO₄ to calculate the mass of Na₂SO₄.
Mass of Na₂SO₄ = 0.375 mol Na₂SO₄ × (142.04 g Na₂SO₄/1 mol Na₂SO₄) = 53 g Na₂SO₄
The reaction produces of Na₂SO₄.
Answer:
Using the drop down Menu
Proton : Possessing a Charge of +1. It is found in the Nucleus and has a Mass of "A"
Neutron; Possesses a charge of 0. It is situated in the Nucleus and Has a Mass of "B"
Electron: Has a charge of -1. Found in the Orbitals of the atom and has a mass of "C"
