Question

If $ 50g $ of $ CaC{O_3} $ is allowed to react with $ 70g $ of $ {H_3}P{O_4} $ . Calculate:
a) Amount of $ C{a_3}{(P{O_4})_2} $ formed
b) Amount of unreacted reagent

Answer 1

Hint: A limiting agent is the reagent which is completely used up to form the product, and thus the reaction cannot be continued without that particular reagent. In the given reaction Calcium carbonate ( $ CaC{O_3} $ ) is the limiting reagent.

Complete step by step solution
In this type of question we have to write a balanced chemical equation. After that the limiting reagent must be checked, so that the driving force of the reaction was easily detected.
From the mentioned question we can write the equation as-
 $ 3CaC{O_3} + 2{H_3}P{O_4} \to C{a_3}{(P{O_4})_2} + 3{H_2}O + 3C{O_2} $
The molar mass of Calcium carbonate can be calculated as $ CaC{O_3} = 40 + 12 + 48 = 100gm $
The molar mass of Phosphoric acid can be calculated as $ {H_3}P{O_4} = 3 \times 1 + 31 + 4 \times 16 = 98gm $
The molar mass of Calcium phosphate can be calculated as
  $ C{a_3}{(P{O_4})_2} = 3 \times 40 + 2 \times (31 + 4 \times 16) \\
  C{a_3}{(P{O_4})_2} = 120 + 2 \times 95 = 310gm \\ $
Now the given mass of Calcium carbonate is $ CaC{O_3} = 50gm $
Given mass of Phosphoric acid is $ {H_3}P{O_4} = 70gm $
As we know moles can be calculated by the formula: $ moles(n) = \dfrac{{G.M}}{{M.M}} $
Therefore, the moles calcium carbonate is $ {n_{CaC{O_3}}} = \dfrac{{50}}{{100}} = 0.5 $
And the moles Phosphoric acid is $ {n_{{H_3}P{O_4}}} = \dfrac{{70}}{{98}} = 0.714 $
Since the number moles of calcium carbonate is lesser than the number of moles of phosphoric acid, we can say calcium carbonate is the limiting agent.
A limiting agent is the reagent which is completely used up to form the product, and thus the reaction cannot be continued without that particular reagent.
(a) $ 3CaC{O_3} + 2{H_3}P{O_4} \to C{a_3}{(P{O_4})_2} + 3{H_2}O + 3C{O_2} $
By stoichiometry, $ 3 $ mole of $ CaC{O_3} $ give $ 1 $ mole of $ C{a_3}{(P{O_4})_2} $
 $ \therefore 1 $ Mole of $ CaC{O_3} $ give $ \dfrac{1}{3} $ mole of $ C{a_3}{(P{O_4})_2} $
 $ \therefore 0.5 $ Mole of $ CaC{O_3} $ give $ \dfrac{1}{3} \times 0.5 $ mole of $ C{a_3}{(P{O_4})_2} = 0.166 $
Hence the amount of $ C{a_3}{(P{O_4})_2} $ is
 $ C{a_3}{(P{O_4})_2} = 0.166mol \times molar mass \\
   \Rightarrow C{a_3}{(P{O_4})_2} = 0.166mol \times 310gm \\
  \therefore C{a_3}{(P{O_4})_2} = 51.46gm \\ $
(B) Secondly, by stoichiometry, $ 3 $ mole of $ CaC{O_3} $ give $ 2 $ mole of $ {H_3}P{O_4} $
 $ \therefore 1 $ Mole of $ CaC{O_3} $ give $ \dfrac{2}{3} $ mole of $ {H_3}P{O_4} $
 $ \therefore 0.5 $ Mole of $ CaC{O_3} $ give $ \dfrac{2}{3} \times 0.5 $ mole of $ {H_3}P{O_4} = \dfrac{1}{3} = 0.333 $
Hence the consumed amount of $ {H_3}P{O_4} $ by $ CaC{O_3} $ is
 $ {H_3}P{O_4} = 0.333mol \times molar mass \\
   \Rightarrow {H_3}P{O_4} = 0.333mol \times 98gm \\
  \therefore {H_3}P{O_4} = 32.634gm \\ $
As the given mass of $ {H_3}P{O_4} = 70gm, $
The unreacted amount will be $ = 70gm - 32.634gm = 37.336gm $
Note
In chemistry laboratories Phosphoric acid ( $ {H_3}P{O_4} $ ) is also known as orthophosphoric acid. Here the oxidation state of Phosphorus is $ + 5 $ . The pure compound is colourless solid. Calcium carbonate ( $ CaC{O_3} $ ) can decompose while heating to give calcium oxide and carbon dioxide.