Answer:
The required mass of oxygen to react with 4 moles of Mg is 64 grams
Explanation:
You know the reaction where magnesium is burned with oxygen to form magnesium oxide:
2Mg + \(O_{2}\) --> 2MgO
Oxygen is required to react completely (that is, sufficient oxygen must be present for the reaction to occur) with 4 moles of Mg. For that you observe the stoichiometry of the reaction, the amount of reagents needed. Stoichiometry in this case tells you that in order for 2 moles of Mg to react, you need 1 mole of \(O_{2}\). Taking this into account, you apply a rule of three: if to react with two moles of Mg you need 1 mole of \(O_{2}\), how many moles of \(O_{2}\) do you need to react with 4 moles of Mg?
\(moles of O_{2} =\frac{4 molesMg*1molO_{2} }{2 molesMg}\)
moles of O_{2} =2 moles
To know the mass of \(O_{2}\) that reacts, you must take into account the molar mass of oxygen, which is 32 grams / mol and you can apply another rule of three: if in a mole of \(O_{2}\) there are 32 grams, how many grams are in 2 moles of oxygen?
\(mass of oxygen=\frac{2 moles* 32 grams}{1mol}\)
mass of oxygen= 64 grams
Then, the required mass of oxygen to react with 4 moles of Mg is 64 grams
