Question

Potassium hydroxide readily reacts With Co2 ro Produce Pottasium Carbonate and Water. How Many Grams of Pottasium Carbonate are Produced If 224.4 g of Pottasium Hydroxode Reacts With 88g Of Co2 The Reaction also Produces 36g Of Water

Chapter Name : Atoms And Molecules
Class : 9 (IX)
Syllabus : NCERT (CBSE)

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Answer 1

Answer:

If $224.4$ g of potassium hydroxide reacted with $88$ g of carbon dioxide, $276$ g of potassium carbonate is obtained as products along with $36$ g of water·

Explanation:

The balanced chemical equation for the given reaction is $:$

    $CO_{2}_(_g_)\ +\ 2KOH_(_a_q_)$   →   $K_{2}CO_{3} _(_a_q_)\ +\ H_{2}O_(_l_)$

From the balanced chemical equation, it is clear that

One molecule of carbon dioxide reacted with two molecules of potassium hydroxide to get one molecule of potassium carbonate and one molecule of water as products·

We know that,

The molecular mass of $CO_{2}$  $=\ \ 44$g

The molecular mass of $KOH$  $=\ \ 56$g

The molecular mass of $K_{2}CO_{3}$  $=\ \ 138$g

The molecular mass of $H_{2}O$  $=\ \ 18$g

We have to find out the weight of $K_{2}CO_{3}$ $($ in grams$)$ when $224.4$ g of potassium hydroxide reacts with $88$ g of carbon dioxide·

Given that,

The weight of $H_{2}O$ formed at that time is $36$ g·

From the molecular mass of $KOH$,

The weight of $2$ moles of $KOH$  is  $=\ 2$ × $56$ g·

                                                          $=\ 112$ g

So from the equation,

   $112$ g of $KOH$ gives $138$ g of $K_{2}CO_{3}$

∴  $224.4$ g of $KOH$ gives   $=$     $\frac{138g \ *\ 224.4g}{112g}$

                                          $=\ \ 276.492$ g

                                          ≈    $276$ g of $K_{2}CO_{3}$

Therefore,

  If $224.4$ g of potassium hydroxide reacted with $88$ g of carbon dioxide, $276$ g of potassium carbonate and $36$ g of water are obtained as products·

∴ the answer is $276$g of $K_{2}CO_{3}$·