1 (a)
Step 1. Write the balanced chemical equation for the reaction
#"Al(OH)"_3 + "3HCl" → "AlCl"_3 + 3"H"_2"O"#
Step 2. Calculate the mass of #"HCl"# produced in one day
#"Mass of HCl" = 2.5 color(red)(cancel(color(black)("L juice"))) × "3.0 g HCl"/(1 color(red)(cancel(color(black)("L juice")))) = "7.5 g HCl"#
Step 3. Calculate the moles of #"HCl"#
#" Moles of HCl" = 7.5 color(red)(cancel(color(black)("g HCl"))) × "1 mol HCl"/(36.46 color(red)(cancel(color(black)("g HCl")))) = "0.206 mol HCl"#
Step 4. Calculate the moles of #"Al(OH)"_3#
#"Moles of Al(OH)"_3 = 0.206 color(red)(cancel(color(black)("mol HCl"))) × ("1 mol Al(OH)"_3)/(3 color(red)(cancel(color(black)("mol HCl")))) = "0.0686 mol Al(OH)"_3#
Step 5. Calculate the mass of #"Al(OH)"_3#
#"Mass of Al(OH)"_3 = 0.0686 color(red)(cancel(color(black)("mol Al(OH)"_3))) × ("78.00 g Al(OH)"_3)/( 1 color(red)(cancel(color(black)("mol Al(OH)"_3)))) = "5.35 g Al(OH)"_3#
Step 6. Calculate the number of antacid tablets
#"Number of tablets" = 5.35 color(red)(cancel(color(black)("g Al(OH)"_3))) × "1 tablet"/(0.400 color(red)(cancel(color(black)("g Al(OH)"_3)))) = "13 tablets"#
1 (b)
Step 1. Write the balanced chemical equation for the reaction
#"Ca(OH"_2 + "2HCl" → "CaCl"_2 +2"H"_2"O"#
Step 2. Calculate the mass of the #"Ca(OH)"_2# solution
#"Mass of solution" = 25 color(red)(cancel(color(black)("cm"^3color(white)(l) "solution"))) × "1.025 g solution"/(1 color(red)(cancel(color(black)("cm"^3 color(white)(l)"solution")))) = "25.6 g solution"#
Step 3. Calculate the mass of #"Ca(OH)"_2#
#"Mass of Ca(OH)"_2 = 25.6 color(red)(cancel(color(black)("g solution"))) × "1.85 g Ca(OH)"_2/(100 color(red)(cancel(color(black)("g solution")))) = "0.474 g Ca(OH)"_2#
Step 4. Calculate the moles of #"Ca(OH)"_2#
#"Moles of Ca(OH)"_2 = 0.474 color(red)(cancel(color(black)("g Ca(OH)"_2))) × "1 mol Ca(OH)"_2/(74.09 color(red)(cancel(color(black)("g Ca(OH)"_2)))) = "0.006 40 mol Ca(OH)"_2#
Step 5. Calculate the moles of #"HCl"#
#"Moles of HCl" = "0.006 40" color(red)(cancel(color(black)("mol Ca(OH)"_2))) × "2 mol HCl"/(1 color(red)(cancel(color(black)("mol Ca(OH)"_2)))) = "0.0128 mol HCl"#
Step 6. Calculate the volume of #"HCl"#
#"Volume of HCl" = 0.0128 color(red)(cancel(color(black)("mol HCl"))) × "1 L HCl"/(0.1 color(red)(cancel(color(black)("mol HCl")))) = "0.13 L HCl" = "130 mL HCl"#
