Question

Calculate mass, volume and number of molecules of MgO when 100 grams of O2 reacts with magnesium

Answer 1

Answer:

moles of MgO formed by 100 g $O_{2}$ = 6.25

Therefore mass of MgO formed = 6.25×40= 250g

Volume of MgO formed = 6.25×22.4 L = 5600 L

No. of molecules of MgO = 6.25 × 6.022 ×$10^{23}$= 3.76 ×$10^{24}$

Explanation:

First of all we write a balanced chemical equation for the given reaction-

2Mg(s) + $O_{2}$ (g) ⇒ 2MgO (g)

From the above reaction we get that 2 moles of Mg gives 2 moles of Mgo

and 1 mole of $O_{2}$ gives 2 moles of MgO

Since product formed per mole of Mg is least, so Mg is the limiting reagent here.

Now, Moles of $O_{2}$ required to react with 2 moles of Mg = 1

Given no. of Moles of $O_{2}$ = 100/32= 3.125

So. moles of Mg required= 2 × 3.125= 6.25

Now, Moles of MgO formed by 1 mole of Mg = 1

Therefore moles of MgO formed by 100 g $O_{2}$ = 6.25

Therefore mass of MgO formed = 6.25×40= 250g

Volume of MgO formed = 6.25×22.4 L = 5600 L

No. of molecules of MgO = 6.25 × 6.022 ×$10^{23}$= 3.76 ×$10^{24}$

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