Question

Question 30

Answer 1

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,

$Mg(ClO_3)_2+2NaOH\rightarrow Mg(OH)_2+2NaClO_3$

From the balanced reaction we conclude that

As, 1 moles of $Mg(ClO_3)_2$ react with 2 mole of $NaOH$

So, 2.72 moles of $Mg(ClO_3)_2$ react with $2\times 2.72=5.44$ moles of $NaOH$

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

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

As, 2 moles of $NaOH$ react to give 1 mole of $Mg(OH)_2$

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

And,

As, 2 moles of $NaOH$ react to give 2 mole of $NaClO_3$

So, 3.14 moles of $NaOH$ react to give 3.14 moles of $NaClO_3$

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

Answer 2

well #1 is A

and #2 is E  

well i did just do this