In an experiment 2.750 g of magnesium is reacted with 2.571 g of oxygen gas according to the equation:
Mg + O2 ==> MgO.
What mass of the other reactant is in exess?
see other post.
To find the mass of the other reactant in excess, we first need to determine the limiting reactant in the given reaction. The limiting reactant is the one that will be completely consumed, and any excess reactant will be left over.
To do this, we need to compare the moles of both reactants.
1. Calculate the moles of magnesium (Mg):
Moles of Mg = Mass of Mg / Molar mass of Mg
Molar mass of Mg = 24.31 g/mol
Moles of Mg = 2.750 g / 24.31 g/mol
2. Calculate the moles of oxygen gas (O2):
Moles of O2 = Mass of O2 / Molar mass of O2
Molar mass of O2 = 32.00 g/mol
Moles of O2 = 2.571 g / 32.00 g/mol
3. Now, compare the moles of Mg and O2 to determine the limiting reactant.
The balanced equation tells us that 1 mole of Mg reacts with 1 mole of O2 to form 1 mole of MgO. Therefore, the mole ratio is 1:1.
In this case:
- Moles of Mg = 0.113 mol
- Moles of O2 = 0.080 mol
Since the mole ratio is 1:1, we can conclude that there is an excess of O2.
4. To find the mass of the excess reactant (O2), we can use the molar mass:
Mass of excess reactant (O2) = Moles of excess reactant (O2) * Molar mass of O2
Mass of excess reactant (O2) = (0.080 mol) * (32.00 g/mol)
Therefore, the mass of the excess reactant is 2.56 g (rounded to two decimal places).
In conclusion, the mass of the other reactant (oxygen gas) that is in excess is 2.56 g.