one gram of magensium is burnt in a closed vessel which contains 0.5 grams of oxygen.
which is limiting reactant?
what is the amount of mgo formed ?
To determine the limiting reactant, we need to compare the number of moles of magnesium (Mg) with the number of moles of oxygen (O2) and see which one is in a stoichiometrically smaller amount.
First, let's calculate the number of moles of magnesium (Mg):
Molar mass of Mg = 24.31 g/mol
Number of moles of Mg = Mass of Mg / Molar mass of Mg
Number of moles of Mg = 1 g / 24.31 g/mol
Number of moles of Mg = 0.041 moles
Next, let's calculate the number of moles of oxygen (O2):
Molar mass of O2 = 32.00 g/mol
Number of moles of O2 = Mass of O2 / Molar mass of O2
Number of moles of O2 = 0.5 g / 32.00 g/mol
Number of moles of O2 = 0.016 moles
From the calculations, we can see that magnesium is in excess as it has more moles (0.041 moles) compared to oxygen (0.016 moles). Therefore, the limiting reactant is oxygen.
Now, let's determine the amount of magnesium oxide (MgO) formed using the limiting reactant (oxygen):
The balanced chemical equation for the reaction between magnesium and oxygen is:
2 Mg + O2 -> 2 MgO
From the stoichiometry, we know that 1 mole of Mg reacts with 0.5 moles of O2 to form 1 mole of MgO.
Since oxygen is the limiting reactant, we use its number of moles (0.016 moles) to calculate the amount of MgO formed.
Number of moles of MgO = 0.016 moles
Finally, to determine the mass of MgO formed, we can use the molar mass of MgO:
Molar mass of MgO = 40.31 g/mol
Mass of MgO formed = Molar mass of MgO x Number of moles of MgO
Mass of MgO formed = 40.31 g/mol x 0.016 moles
Mass of MgO formed = 0.645 g
Therefore, the amount of MgO formed is 0.645 grams.
2Mg+O2>>2MgO
so how many moles of Mg do you have?
moles Mg=1/24.3
then according to the balanced equation, you need 1/2 * 1/24.3 moles O2. WHat is the mass of that? 1/2*1/24.3*32 grams, which in my head is more than .5 grams, so you are limited by the O2.