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. – 6145

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1. Work out the Mr (relative formula mass) of copper chloride (CuCl₂):
   Cu = 63.5, Cl = 35.5 × 2
   Mr CuCl₂ = 63.5 + (2 × 35.5) = 134.5
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   Calculate the number of moles of CuCl₂ the student wants to make:
   moles = mass ÷ Mr = 11.0 ÷ 134.5 ≈ 0.0818 mol
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   Use the mole ratio from the balanced equation (1 mole CuCO₃ → 1 mole CuCl₂), so moles of CuCO₃ needed = 0.0818 mol.

   Find the Mr of copper carbonate (CuCO₃):
   Cu = 63.5, C = 12, O = 16 × 3
   Mr CuCO₃ = 63.5 + 12 + (3 × 16) = 123.5
   Turton School
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   Calculate the mass of CuCO₃ required:
   mass = moles × Mr = 0.0818 × 123.5 ≈ 10.1 g

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