Question

According to the following reaction, what amount of Al2S3 remains when 30.00g of Al2S3 and 4.00g of H2O are reacted?

A few of the molar masses are as follows: Al2S3=150.17 g/mol
H2O=18.02 g/mol

Al2S3(s)+ 6 H2O(l) -> 2Al (OH)3(s)+3H2S(g)

Answer 1

The given reaction is Al_2S_3(s) + 6H_2O(1) rightarrow 2Al(OH)_3 (s) + 3H_2S(g) moles of Al_2S_3 = mass /molar mass = 30.00g/150.17 g mol^-1 = 0.199 moles and moles of H_2O = 4.00g/18.02g mol^-1 = 0.222 moles According to the equation , 1 mole of Al_2S_3 reacts with 6 moles of H_2O molecules Hence for 0.222 moles of H_2O molecules, Al_2S_3 required = 0.222/6 = 0.037 moles But Al_2S_3 used = 0.199 moles Hence moles of Al_2S_3 remains = 0.199-0.037 = 0.162 moles Now, 0.162 moles of Al_2S_3 = 0.162 mol times 150.17 g mol^-1 = 24.33 g of Al_2S_3