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Question 30

Select the correct answer from each drop-down menu.


A reaction proceeds with 2.72 moles of magnesium chlorate and 3.14 moles of sodium hydroxide. This is the equation of the reaction:

Mg(ClO3)2 + 2NaOH → Mg(OH)2 + 2NaClO3.


Determine the theoretical amount of each product that the reaction will produce.


The reaction will produce 1.____ of magnesium hydroxide and 2.____ of sodium chlorate.



(Both have the same dropdown)

A. 1.36

B. 1.57

C. 2.72

D. 3.14

E. 4.52

(All in moles)

Respuesta :

well #1 is A

and #2 is E  

well i did just do this

Answer : The reaction will produce (B) 1.57 mole of magnesium hydroxide and (D) 3.14 mole of sodium chlorate.

Explanation :

First we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

[tex]Mg(ClO_3)_2+2NaOH\rightarrow Mg(OH)_2+2NaClO_3[/tex]

From the balanced reaction we conclude that

As, 1 moles of [tex]Mg(ClO_3)_2[/tex] react with 2 mole of [tex]NaOH[/tex]

So, 2.72 moles of [tex]Mg(ClO_3)_2[/tex] react with [tex]2\times 2.72=5.44[/tex] moles of [tex]NaOH[/tex]

That means, in the given balanced reaction, [tex]NaOH[/tex] is a limiting reagent because it limits the formation of products and [tex]Mg(ClO_3)_2[/tex] is an excess reagent.

Now we have to calculate the moles of [tex]Mg(OH)_2[/tex] and [tex]NaClO_3[/tex] by using the moles of limiting reagent.

As, 2 moles of [tex]NaOH[/tex] react to give 1 mole of [tex]Mg(OH)_2[/tex]

So, 3.14 moles of [tex]NaOH[/tex] react to give [tex]\frac{3.14}{2}=1.57[/tex] moles of [tex]Mg(OH)_2[/tex]

And,

As, 2 moles of [tex]NaOH[/tex] react to give 2 mole of [tex]NaClO_3[/tex]

So, 3.14 moles of [tex]NaOH[/tex] react to give 3.14 moles of [tex]NaClO_3[/tex]

Therefore, the reaction will produce (B) 1.57 mole of magnesium hydroxide and (D) 3.14 mole of sodium chlorate.

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