Determine how many grams of active antacid ingredient AL(OH)3 would be necessary to neutralize the HCL in 50.0 mL of .0100 M HCL (an approximation of the conc. of HCL in the stomach)
3HCL9aq)+AL(OH)3(s)-> AlCL3 (aq) = 3H2O (L)
stop it. YOu are posting questions with no work, looking for someone to do your thinking and work for you.
20
To determine the number of grams of active antacid ingredient AL(OH)3 required to neutralize the HCl in the given solution, we need to use stoichiometry.
First, let's write the balanced chemical equation for the reaction:
3HCl(aq) + Al(OH)3(s) -> AlCl3(aq) + 3H2O(l)
From the equation, we can see that 3 moles of HCl react with 1 mole of Al(OH)3.
To find the moles of HCl in the solution, we need to use the equation:
moles = concentration x volume
moles of HCl = 0.0100 M x 0.0500 L = 0.0005 moles
Since the stoichiometric ratio between HCl and Al(OH)3 is 3:1, we can calculate the moles of Al(OH)3 required as follows:
moles of Al(OH)3 = (0.0005 moles of HCl) / (3 moles of HCl / 1 mole of Al(OH)3)
moles of Al(OH)3 = 0.0005 moles / 3
moles of Al(OH)3 = 0.000167 moles
Finally, we can calculate the grams of Al(OH)3 required using its molar mass:
grams = moles x molar mass
grams of Al(OH)3 = 0.000167 moles x (26.98 g/mol for Al + (3 x 16.00 g/mol) for OH + (3 x 1.01 g/mol) for H)
grams of Al(OH)3 = 0.000167 moles x 78.01 g/mol
grams of Al(OH)3 = 0.013 g (rounded to three decimal places)
Therefore, approximately 0.013 grams of active antacid ingredient AL(OH)3 would be necessary to neutralize the HCl in 50.0 mL of 0.0100 M HCl solution.