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2Mg + O2 → 2Mgo

What is the greatest amount of MgO (in moles) that can be made with 7.8 moles of Mg and 4.7
moles of O,? Which is the limiting reactant? Which reactant is in excess, and how many moles
of it are left over?

Sagot :

s924950

Answer:

it would be 4.5 mole left over

Explanation:

The greatest amount of MgO will be 7.8 mole and Mg will be limiting reagent.

What is limiting reagent?

The substance that would be entirely absorbed in the end of the chemical process is called as the limiting reagent.

What is mole?

A mole would be defined as the quantity of a material that includes precisely 6.02214076 X [tex]10^{23}[/tex] of the particular chemical elementary entities.

Calculation of amount of  MgO

The balanced chemical equation is 2 Mg + [tex]O_{2}[/tex] → 2MgO

It can be seen that 2 mole of Mg reacts with 1 mole of oxygen.

So, 7.8 mole of Mg = 7.8 / 2 = 3.9 mole of [tex]O_{2}[/tex].

Since, [tex]O_{2}[/tex] will be added in given reaction vessel will be 4.7 moles.

So, (4.7 - 3.9) = 0.8 moles of [tex]O_{2}[/tex] left but in this case Mg will be finished. Therefore, Mg will be limiting reagent.

It can be understood that 2 moles of Mg will give 2 mole of MgO.

That's why , 7.8 moles of Mg = 2 × 7.8 / 2 = 7.8 moles of MgO.

Therefore, 7.8 mole of magnesium oxide (MgO) could be generated and Mg will be limiting reagent.

To know more about limiting reagent

https://brainly.com/question/11848702

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