Question

2grm of mixture of caco3 and mgco3 required 2grm of H2SO4 to complete reaction. determine the percentage composition of mixture ?​

Answer 1

Given :

Amount of mixture of CaCO₃ and MgCO₃ = 2g

Amount of H₂SO₄ = 2g

To Find :

The percentage composition of the mixture

Solution :

• The following reactions will take place

           $MgCO_{3} \rightarrow MgO+CO_{2} \\ CaCO_{3} \rightarrow CaO+CO_{2}$

• Molecular mass of MgCO₃ = 84gm

        Molecular mass of CaCO₃ = 100gm

        Molecular mass of H₂SO₄ = 98gm

• Number of moles of acid   $=\frac{2}{98} =0.02$

• 0.01 moles reacts with MgCO₃ and 0.01 moles react with CaCO₃

• Percentage of MgCO₃ = $\frac{0.84\times100}{2} =42\%$

        Percentage of CaCO₃ =  $\frac{98\times100}{2} =58\%$

The percentage composition of the mixture is MgCO₃(42%) and CaCO₃(58%)

Answer 2

Answer:

Amount of mixture of CaCO₃ and MgCO₃ = 2g

Amount of H₂SO₄ = 2g

To Find :

The percentage composition of the mixture

Solution :

   The following reactions will take place

         

   Molecular mass of MgCO₃ = 84gm

       Molecular mass of CaCO₃ = 100gm

       Molecular mass of H₂SO₄ = 98gm

   Number of moles of acid  

   0.01 moles reacts with MgCO₃ and 0.01 moles react with CaCO₃

   Percentage of MgCO₃ =

       Percentage of CaCO₃ =  

The percentage composition of the mixture is MgCO₃(42%) and CaCO₃(58%)

Explanation: