If 3.00g of aluminum hydroxide reacts with 1.00g of sulfuric acid according to the balanced equation, what is the mass of water produced?Al(OH)3 (s) + H2 SO4 (l) -gt; Al2 (SO4)3 (aq) + HOH (l)the answer is 8.367g H2O Limited reactant (LR) is H2SO4How do you do this problem???
first you have to convert 3.00g of Al(OH)3 to the number of moles of Al(OH)3to do that you have to find the molecular mass of Al(OH)3 by adding aluminun (26.98 g/mol) to three of OH, which is 16+117 .times 3 51 g/mol and add that to aluminun which you should get 77.98 g/molto find the number of moles in 3.00g of Al(OH)3, you simply have to do 3.00g / 77.98 g/mol, getting .038 moles of Al(OH)3you must do the same for sulfuric acid with the 1.00g, you should get for the molecular mass 98.1g/moland divide that from 1.00g to get0.0102 moles of sulfuric acidand you should known by the coefficients of Al(OH)3 and H2 SO4, that ONE MOLE of Al(OH)3 added to ONE MOLE of the acid creates the water and stuff so .0102 MOLES OF H2SO4 WILL MATCH .0102 MOLES OF ALUMINUM HYDROXIDE TO FORM THE WATERbut there is still 0.0278 moles of the hydroxide left, making the acid the limited reactantso you will also tell by the coefficients that every mole of the hydroxide OR the acid will equal the moles of water(one mole of hydroxide + one mole of acid one mole of water) so there is 0.0102 moles of H20, or HOHand you should know that the molecular mass of water is 18.00 g/molso do the opposite of what we did in the beginning to find the number of moles and multiply the molecula mass of water by the moles to leave you with the number of grams, which isOH DANG IT!!! THAT EQUATION ISN'T FREAKING BALANCED AT ALL!!!! i guess you'll have to balance the equation first before you do all the stuff i just said.
