Question

How many grams of cobalt(III) chloride are needed to react completely with $ 68.9mL $ of $ 366M $ $ KOH $ solution, if the equation for the reaction is $ C{o^{2 + }}_{(aq)} + 2O{H^ - }_{(aq)} \to Co{(OH)_2}(8) $ ?

Answer 1

Hint : To find the required value we will use the concept of molarity, now molarity is defined for the amount of a substance in a certain volume of solution as moles of solute per litre of solution. It is a unit of measurement of concentration.

Complete Step By Step Answer:
Let's try to find the number of moles of $ KOH $ that is used in this reaction, once we get to know that we can use the mole relationship which we get through our given reaction
So as we know molarity
 $ c = \dfrac{n}{V} $ , where $ n = $ moles of solute, $ V = $ volume of solution
So by using this formula we know moles of $ KOH $ will be equal to
 $
  n = c.V \\
   = 0.366M \times 68.9 \times {10^{ - 3}}L \\
   = 0.02522moles \\
  $
Hence, moles of $ KOH $ is $ 0.02522 $
So the relationship through the reaction is one mole of $ CoC{l_2} $ is required to neutralise $ 2mol $ $ KOH $
And by using this relation,
 $
  0.02522KOH \times \dfrac{{1.CoC{L_2}}}{{2.KOH}} \\
   = 0.01261molesCoC{l_2} \\
  $
So now we can determine how many grams would contain this moles
 $
  0.01261moles \times \dfrac{{129.839g}}{{1mole}} \\
   = 1.64gCoC{l_2} \\
  $

Note :
To find the concentration of a given solution there are various ways to find it. There are different concepts such as molality, normality and as we have seen in above solution molarity to find out concentration. As per the requirements and as per availability of given data we decide which method to prefer.