Exercise Aluminum Hydroxide (Al(OH)_(3)) is used as an antiacid in some pharmaceutical formulations. It reacts with a typical acid HCl according to the equation: Al(OH)_(3)(s)+3HCl(aq)arrow AlCl_(3)((aq)+3H_(2)O(l) How many grams of HCl are need to neutralize 5.00 g HCl? (MAl(OH)_(3)=78.00g/mol,M_(CO)=36.5g/mol

To find out how many grams of HCl are needed to neutralize 5.00 g of Al(OH)3, we need to use the balanced chemical equation and the molar masses of the reactants.<br /><br />The balanced chemical equation is:<br />Al(OH)3(s) + 3HCl(aq) → AlCl3(aq) + 3H2O(l)<br /><br />From the equation, we can see that 1 mole of Al(OH)3 reacts with 3 moles of HCl to produce 1 mole of AlCl3 and 3 moles of H2O.<br /><br />Given:<br />Molar mass of Al(OH)3 = 78.00 g/mol<br />Molar mass of HCl = 36.5 g/mol<br /><br />To find the amount of HCl needed, we need to calculate the moles of Al(OH)3 and then use the stoichiometry of the reaction to find the moles of HCl required.<br /><br />Moles of Al(OH)3 = 5.00 g / 78.00 g/mol = 0.0641 mol<br /><br />Using the stoichiometry of the reaction, we can see that 1 mole of Al(OH)3 reacts with 3 moles of HCl. Therefore, 0.0641 mol of Al(OH)3 will react with 0.0641 mol * 3 = 0.1923 mol of HCl.<br /><br />Now, we can calculate the mass of HCl required:<br /><br />Mass of HCl = 0.1923 mol * 36.5 g/mol = 7.00 g<br /><br />Therefore, 7.00 grams of HCl are needed to neutralize 5.00 grams of Al(OH)3.