For the following question(s), consider the following equation.
2Mg + 02 → 2Mgo
How many grams of MgO are produced when 40.0 grams of 02 react completely with Mg?
60.5 g
202 g
50.4 g
40.3 g
101 g
To determine the number of grams of MgO produced, we need to find the molar mass of MgO.
The molar mass of Mg is approximately 24.31 g/mol.
The molar mass of O is approximately 16.00 g/mol.
In the balanced equation, we see that 2 moles of Mg reacts to form 2 moles of MgO.
So, 1 mole of Mg reacts to form 1 mole of MgO.
Therefore, the molar mass of MgO is approximately:
(24.31 g/mol) + (16.00 g/mol) = 40.31 g/mol
Now, we can calculate the number of moles of O2 present in 40.0 grams using its molar mass:
40.0 g / (32.00 g/mol) = 1.25 mol
Since 1 mole of Mg reacts with 1 mole of MgO, the number of moles of MgO produced is also 1.25 mol.
Finally, we can calculate the mass of MgO produced using its molar mass:
1.25 mol * (40.31 g/mol) = 50.39 g
Therefore, the correct answer is 50.4 g.
To find the number of grams of MgO produced, we need to use the stoichiometry of the balanced chemical equation.
The balanced equation is: 2Mg + O2 → 2MgO
From the equation, we can see that 2 moles of Mg react with 1 mole of O2 to produce 2 moles of MgO.
First, let's convert the given mass of O2 to moles using the molar mass of O2:
Molar mass of O2 = 16.00 g/mol + 16.00 g/mol = 32.00 g/mol
Number of moles of O2 = 40.0 g / 32.00 g/mol ≈ 1.25 mol (rounded to 2 decimal places)
According to the stoichiometry of the equation, 2 moles of Mg react with 1 mole of O2 to produce 2 moles of MgO.
So, the number of moles of MgO produced will be the same as the number of moles of Mg.
Now, let's calculate the molar mass of Mg:
Molar mass of Mg = 24.31 g/mol
Number of moles of Mg = 1.25 mol
Mass of MgO produced = Number of moles of MgO × Molar mass of MgO
Mass of MgO produced = 1.25 mol × (2 moles MgO / 2 moles Mg) × (24.31 g/mol MgO)
Mass of MgO produced = 1.25 mol × 24.31 g/mol MgO
Mass of MgO produced ≈ 30.3875 g
Rounding to the nearest tenth, we have:
Mass of MgO produced ≈ 30.4 g
Therefore, the closest answer choice is 30.4 g, which is not one of the given options.
To calculate the mass of MgO produced when 40.0 grams of O2 react completely with Mg, we first need to find the molar mass of MgO.
The molar mass of an element is the mass of one mole of that element. Since Mg and O are the elements in MgO, we need to find their respective molar masses.
The molar mass of Mg is 24.31 g/mol since it has an atomic mass of 24.31 g/mol.
The molar mass of O is 16.00 g/mol since it has an atomic mass of 16.00 g/mol.
Next, we need to calculate the number of moles of O2. To do this, we divide the mass of O2 by its molar mass.
Given:
Mass of O2 = 40.0 grams
Molar mass of O2 = 32.00 g/mol (16.00 g/mol x 2)
Number of moles of O2 = Mass of O2 / Molar mass of O2
= 40.0 g / 32.00 g/mol
= 1.25 mol
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, since the ratio is 2 moles of MgO to 1 mole of O2, the number of moles of MgO formed will be half of the moles of O2.
Number of moles of MgO = 1.25 mol / 2
= 0.625 mol
Finally, we can calculate the mass of MgO by multiplying the number of moles by the molar mass of MgO.
Mass of MgO = Number of moles of MgO x Molar mass of MgO
= 0.625 mol x (24.31 g/mol + 16.00 g/mol)
= 0.625 mol x 40.31 g/mol
= 25.19 g
Therefore, the answer is approximately 25.19 grams of MgO. None of the given options match this value.