What mass of water is formed by the interaction of 9.8 g of copper hydroxide with carbonic acid?

1. The interaction of copper hydroxide with carbonic acid is described by the reaction equation:

Cu (OH) 2 + H2CO3 = CuCO3 + 2H2O;

2.Calculate the chemical amount of copper hydroxide:

n (Cu (OH) 2) = m (Cu (OH) 2): M (Cu (OH) 2);

M (Cu (OH) 2) = 64 + 2 * 17 = 98 g / mol;

n (Cu (OH) 2) = 9.8: 98 = 0.1 mol;

3. Determine the amount of water formed:

n (H2O) = n (Cu (OH) 2) * 2 = 0.1 * 2 = 0.2 mol;

4.Calculate the mass of water:

m (H2O) = n (H2O) * M (H2O) = 0.2 * 18 = 3.6 g.

Answer: 3.6 g.




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