If 48 g of magnesium reacts with oxygen gas, how many grams of magnesium oxide will be
formed according to the following equation?
2 Mg + O2 → 2 Mgo

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.

i think the answer is b.

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