how do i write a balanced equation for 6.5g of magnesium burnt in oxygen?

Asked on by stev27197

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sanjeetmanna | College Teacher | (Level 3) Assistant Educator

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First we should write the reaction then we will try to balance the reaction..

This is the reaction between Magnesium burned in oxygen.

Mg + O2 → MgO

Balance the reaction

2Mg + O2 → 2MgO

But given is 6.5 gm of Magnesium...

Now We have to find the moles of 6.5 gm of Magnesium and then using that moles we have to balance the reaction.

Moles = mass/molar mass

         = 6.5/24.3

         = 0.26 moles.

So 0.26 moles of Magnesium is burned in air so what is the reaction...

0.26 Mg + O2 → 0.26 MgO

Now we should know the moles of O2 reacting with 0.26moles of Mg

Moles of O2 when 0.26 moles of Mg is reacting = 0.26/2 = 0.13

So the final reaction will be 

0.26 Mg + 0.13 O2 → 0.26MgO.

So from the reaction we can say that 0.26 moles of Mg when burned in air it will give 0.26 moles of MgO 

since O2 is in excess so Mg will be the limiting reactant and we always know the amount of product formed depends upon the amount of limiting reactant.

Note :- Solution to prove O2 is in excess just by finding the number of molecules of O2

Molecules of O2 = 0.13 * 6.023 * 10^23(N=Avogadro number)

                       = 7.82 * 10^22 Too excess in amount thus proves that O2 is in excess.

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