Using the balalnced chemical equation, how many grams of MgO are produced when 31.0 grams of O2 react completely with Mg?
2 Mg + O2 → 2 MgO
Molar Mass
Mg = 24.3 g/mol
O2 = 32.0 g/mol
MgO = 40.3 g/mol
55.7
78.1
89.1
111.3
To solve this problem, we need to use the stoichiometry of the balanced chemical equation and convert the mass of oxygen (O2) to the mass of magnesium oxide (MgO).
First, calculate the moles of oxygen (O2) using the given mass and molar mass:
moles of O2 = mass of O2 / molar mass of O2
moles of O2 = 31.0 g / 32.0 g/mol
moles of O2 = 0.969 mol
Next, use the balanced chemical equation to determine the ratio between moles of oxygen (O2) and moles of magnesium oxide (MgO):
2 moles of MgO = 1 mole of O2
Now, calculate the moles of magnesium oxide (MgO) produced using the mole ratio:
moles of MgO = 0.969 mol O2 x (2 moles MgO / 1 mole O2)
moles of MgO = 1.939 mol
Finally, convert the moles of magnesium oxide (MgO) to grams using the molar mass:
mass of MgO = moles of MgO x molar mass of MgO
mass of MgO = 1.939 mol x 40.3 g/mol
mass of MgO = 78.2 g
Therefore, when 31.0 grams of oxygen (O2) react completely with magnesium (Mg), 78.2 grams of magnesium oxide (MgO) are produced.
So the correct answer is: 78.1 grams