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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
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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.The balanced chemical equation is:Al(OH)3(s) + 3HCl(aq) → AlCl3(aq) + 3H2O(l)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.Given:Molar mass of Al(OH)3 = 78.00 g/molMolar mass of HCl = 36.5 g/molTo 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.Moles of Al(OH)3 = 5.00 g / 78.00 g/mol = 0.0641 molUsing 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.Now, we can calculate the mass of HCl required:Mass of HCl = 0.1923 mol * 36.5 g/mol = 7.00 gTherefore, 7.00 grams of HCl are needed to neutralize 5.00 grams of Al(OH)3.