A student wanted to make 11.0 g of copper chloride. The equation for the reaction is: CuCO3 + 2HCl → CuCl2 + H2O + CO2 Relative atomic masses, Ar: H = 1; C = 12; O = 16; Cl = 35.5; Cu = 63.5 Calculate the mass of copper carbonate the student should react with dilute hydrochloric acid to make 11.0 g of copper chloride. – 6110

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Q1.

A student wanted to make 11.0 g of copper chloride. 

The equation for the reaction is: 

CuCO3 + 2HCl → CuCl2 + H2O + CO2 

Relative atomic masses, Ar: H = 1; C = 12; O = 16; Cl = 35.5; Cu = 63.5 

Calculate the mass of copper carbonate the student should react with dilute hydrochloric acid to make 11.0 g of copper chloride. ___________________________________________________________ ___________________________________________________________ ___________________________________________________________ ___________________________________________________________ 

Mass of copper carbonate = _________________________ g (4)

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One thought on “A student wanted to make 11.0 g of copper chloride. The equation for the reaction is: CuCO3 + 2HCl → CuCl2 + H2O + CO2 Relative atomic masses, Ar: H = 1; C = 12; O = 16; Cl = 35.5; Cu = 63.5 Calculate the mass of copper carbonate the student should react with dilute hydrochloric acid to make 11.0 g of copper chloride. – 6110”

  1. Mr of CuCl₂, is 134.5
    moles of copper chloride:
    Moles=mass/𝑀𝑟
    =11/134.5=0.08178

    Mr of copper carbonate (CuCO₃) is 123.5.
    mass of CuCO₃:Moles=mass*𝑀𝑟
    ​
    =0.08178×123.5=10.1g

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