In the reaction- 2 Mg - O 2- 2 MgOWhich of the following is true when 24 g of Mg and 32 g of O 2 react-A- All the oxygen will be consumedB- 1 mol of MgO will be producedC- 0-5 mol of Mg will remain unreactedD- All the above

The correct option is B 1 mol of nbsp; MgO will be producedWe know, molar mass of Mg = 24 g ∴ 24 g Mg = 2424 mol of Mg = 1 mol Again molar mass of of O_2 = 3 ...

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Standard XII

Chemistry

Haber Bosch Process

In the reacti...

Question

In the reaction,
$2 M g + O_{2} \to 2 M g O$
Which of the following is true when 24 g of $M g$ and 32 g of $O_{2}$ react?

All the oxygen will be consumed

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1

mol of

M g O

will be produced

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0.5

mol of

M g

will remain unreacted

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All the above

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Solution

The correct option is B $1$ mol of $M g O$ will be produced
We know, molar mass of $M g$ = 24 g
$∴$ 24 g $M g$ = $\frac{24}{24}$ mol of $M g = 1 m o l$
Again molar mass of of $O_{2}$ = 32 g
$∴$ 32 g $O_{2}$ = $\frac{32}{32}$ mol of $O_{2} = 1 m o l$
Given:
$2 M g + O_{2} \to 2 M g O$
2 moles of $M g$ react with 1 mole of $O_{2}$
So, 1 mole of $M g$ will react with 0.5 moles of $O_{2}$
Hence, $M g$ is the limiting reagent here.

From the reaction, 2 moles $M g$ gives 2 moles of $M g O$
So, 1 mole of $M g$ will give 1 mole of $M g O$

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In the reaction, 2Mg+O2→2MgO Which of the following is true when 24 g of Mg and 32 g of O2 react?

In the reaction,
$2 M g + O_{2} \to 2 M g O$
Which of the following is true when 24 g of $M g$ and 32 g of $O_{2}$ react?