Question
Question: If you started with 7.461 grams of magnesium metal, how much oxygen would be consumed during combust...
If you started with 7.461 grams of magnesium metal, how much oxygen would be consumed during combustion?
Solution
The reaction involved in the above question will be: 2Mg+O2→2MgO. The formulas that can be used are Moles=Molecular massGiven mass, Moles of magnesium x Moles of magnesium in reactionMoles of oxygen, and take the molecular mass of oxygen not atomic mass of the oxygen.
Complete answer:
The question says that there is the combustion of magnesium takes place, so the reaction will be:
2Mg+O2→2MgO
We can see from the equation that 2 moles of magnesium and 1 mole of oxygen produces 2 moles of magnesium oxide. The molar ratio can be written as either 1 mol O22 mol Mg or 2 mol Mg1 mol O2.
We know the atomic mass of magnesium is 24.305 g /mol and the molecular mass of oxygen will be:
2 x 15.99 g /mol = 31.998 g /mol
The value of the given mass of magnesium taken at the starting of the reaction is 7.461 grams, then we can find the number of moles easily by using the formula:
Moles=Molecular massGiven mass
Moles=24.307.461=0.3069736
So, the moles of magnesium taken in the reaction are 0.3069736 moles. With this, we can calculate the moles of oxygen used by using the formula:
Moles of magnesium x Moles of magnesium in reactionMoles of oxygen
Putting the values, we can write:
0.3069736 x21 = 0.15348
So, the moles of oxygen used in the combustion will be 0.15348 moles.
Since this value is in moles, we have to find the grams of oxygen used. We can find the mass from the moles by multiplying the molecular mass with moles.
0.15348 x 31.998 = 4.911 g
Therefore, the amount of oxygen consumed will be 4.911 grams.
Note:
Don’t take the mass of oxygen as 16 because in the reaction molecule of oxygen is taken not the atomic form of oxygen. You have to find the value of oxygen in grams because the value of magnesium in question is taken in grams.