Suppose 411 g aluminum sulfide react with 467 g water.Al2S3 + H2O -gt; Al(OH)3 + H2S unbalanced! What mass of the excess reactant remains?
step 1 - get the equation balanced Al2S3 + 6 H2O - 2 Al(OH)3 + 3 H2S step 2 - find the limiting reagent and determine the reaction mass since the aluminum sulfide and water reacts at a 1:6 ratio 467g H2O 1 part Al2S3 / 6 part H2O 77.83333g so only 77.83333g reactstep 3 - solve therefore it leaves411g Al2S3 - 77.83333g Al2S3 333.16667g Using signifigant figures, the mass you are left with is 333g Al2S3
