## Chemistry: The Central Science (13th Edition)

0,855 g AlCl$_3$ 0,346 g H2O
First, we will find the molar mass of AlCl$_3$: MM(AlCl3) = 78g/mol In 0,500 g of Al(OH)$_3$ there are: $\frac{0,500 g}{78g/mol} = 6,41\times10^{-3} mol AlCl_3$ $6,41\times10^{-3} mol AlCl_3$ reacts with HCl The molar mass of HCl is 133,35 g/mol. The mass of $6,41\times10^{-3}$ mols=$6,41\times10^{-3}$ mols x 133,35 g/mol = 0,855 g HCl Second, we will find mass of water: $6,41\times10^{-3} mol AlCl_3$ reacts with 3H$_2$O: $6,41\times10^{-3} \times3$= 0,0192 mols. The molar mass of H$_2$O is 18 g/mol The mass of 0,0192 mols of water is: 0,0192 x 18 = 0,346 g H$_2$O