79.391 grams of MgO is formed when 48 gms of magnesium reacts with oxygen gas.
Explanation:
Balanced equation to show the production of MgO from Mg and oxygen.
2 Mg + O2 → 2 Mgo
From the equation it can be concluded that 2 moles of Mg is required for the formation of 2 moles of MgO.
now, the given weight of Mg is 48 grams
atomic weight of Mg= 24.3 gram/mole
We can calculate the number of moles in 48 grams of Mg by applying the following formula:
number of moles= mass of the substance or element given÷ mass of the one mole of the element.
Putting the values in the above equation
n= 48÷24.3
= 1.97 moles of Mg
Applying the stoichiometry it can be found that
2 moles of Mg gave 2 moles of MgO
1.97 moles of Mg gave x moles of MgO.
Solving for x
2÷2=1.97÷x
x= 1.97
Thus 1.97 moles of MgO is formed, to know the weight in grams we can use the formula used above as
mass = n × atomic mass
= 1.97 × 40.30 grams/mole
= 79.391 grams of MgO is formed.
