Calculate the mass of H20 produced by the decomposition of 2.21 g of CuC03·Cu(OH)2.

The answer I got was .16 g, and I am pretty sure I got it wrong. So could please show me how to do it?
Thanks

Balanced equation: CuCO3.Cu(OH)2(s) -> 2 CuO(s) + CO2(g) + H2O(l)


CuCO3·Cu(OH)2 is also known as Malachite

Given g Malachite x (1 mol Malachite/molar mass Malachite) x (molar ratio: mol H2O/mol Malachite) x (molar mass H2O/1 mol H2O)


Molar mass CuCO3·Cu(OH)2: 221.12 g/mol

Molar mass H2O: 18.016 g/mol


2.21 g CuCO3·Cu(OH)2 x (1 mol Malachite/221.12 g Malachite) x (1 mol H2O/1 mol Malachite) x (18.016 g H2O/1 mol H2O) = 0.1801 g H2O


Sig figs.... 0.180 g H2O



***It seems you are off by 0.02 g. Maybe the problem occurred with the molar mass of the two substances?