Question

When 8g of oxygen reacts with magnesium then the amount of mgo is formed

Answer 1

Explanation:

Mg + O2 --> 2MgO......................(1)

Here, 1 Mole of Mg reacts with 2 Mole of Oxygen to form 2 Moles of MgO

so, according to the question, there is 8 grams of Mg and O

=> Moles of Mg is 8/24 = 1/3 moles.

=> Moles of Oxygen = 8/32 = 1/4 moles. 

This means Oxygen is limited in comparision with Mg

Now, by equation (1) 

we know that 2 moles of oxygen form 2 moles of MgO

=> => 1 Mole of Oxygen [I am not talking of Oxygen gas] will form 1 mole of MgO

=> 1/4 mole of oxygen atoms will form 1/4 mole of MgO

1 Mole of MgO weighs 56 grams

=> 1/4 mole of MgO weighs 56/4= 14 grams, 

I did the question quickly, so see if there's error, and also, you need to know what is limiting factor. It is the compound/element, which is limited in the reaction, and thus limits the reaction. For example in the above reaction, Mg was in more amount than Oxygen [1/3  >  1/4] so, O is the limiting factor instead of Mg. 

Read more on Brainly.in - https://brainly.in/question/11108#readmore

Answer 2

Answer:

When 8g of oxygen reacts with magnesium then the amount of MgO formed is 20 grams.

Explanation:

The reaction of Magnesium with Oxygen proceeds as follows:-

2Mg+O₂⇒2MgO                    (1)

Where,

Mg=Magnesium

O₂=Oxygen

MgO= magnesium oxide

The molecular weight of Mg=24 grams

The molecular weight of O₂=32 grams

The molecular weight of MgO=40

From equation (1) we can see that from 32 grams of oxygen 80 grams of MgO is produced. So, from 1 gram of oxygen, the amount of MgO produced is,

$\frac{80}{32}$            (2)

Now, 8g of oxygen will produce,

$\mathrm{\frac{80}{32}\times 8=20 \ grams}$       (3)

When 8g of oxygen reacts with magnesium then the amount of MgO formed is 20 grams.

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