If 0.81 grams of magnesium oxide are formed, how many grams of oxygen reacted
2Mg + O2 = 2MgO
so, how many moles of Mg in 0.81g ?
You'll need half that many moles of O2
how many grams is that?
To find the mass of oxygen that reacted, we need to use the stoichiometry of the reaction. The balanced chemical equation for the formation of magnesium oxide (MgO) is:
2 Mg + O2 --> 2 MgO
From the balanced equation, we can see that for every 1 mole of magnesium oxide, 1 mole of oxygen molecules (O2) reacts. We need to convert the given mass of magnesium oxide into moles using the molar mass of MgO.
The molar mass of magnesium oxide (MgO) is calculated by adding the atomic masses of magnesium (Mg) and oxygen (O):
Mg: 24.31 g/mol
O: 16.00 g/mol
MgO: 24.31 + 16.00 = 40.31 g/mol
Now, let's convert the given mass of magnesium oxide into moles:
0.81 g MgO * (1 mol MgO / 40.31 g MgO) = 0.0201 mol MgO
Since the stoichiometric ratio tells us that 2 moles of MgO react with 1 mole of O2, we can determine the moles of oxygen that reacted by dividing the moles of MgO by 2:
0.0201 mol MgO * (1 mol O2 / 2 mol MgO) = 0.010 mol O2
Now, to find the mass of oxygen that reacted, we multiply the moles of oxygen by the molar mass of oxygen (O):
0.010 mol O2 * 32.00 g/mol = 0.32 g O2
Therefore, 0.32 grams of oxygen reacted in the formation of 0.81 grams of magnesium oxide.