# Question #5bc2b

Jul 6, 2017

WARNING! Long answer! (a) 13 tablets; (b) 130 mL.

#### Explanation:

1 (a)

Step 1. Write the balanced chemical equation for the reaction

$\text{Al(OH)"_3 + "3HCl" → "AlCl"_3 + 3"H"_2"O}$

Step 2. Calculate the mass of $\text{HCl}$ produced in one day

$\text{Mass of HCl" = 2.5 color(red)(cancel(color(black)("L juice"))) × "3.0 g HCl"/(1 color(red)(cancel(color(black)("L juice")))) = "7.5 g HCl}$

Step 3. Calculate the moles of $\text{HCl}$

$\text{ Moles of HCl" = 7.5 color(red)(cancel(color(black)("g HCl"))) × "1 mol HCl"/(36.46 color(red)(cancel(color(black)("g HCl")))) = "0.206 mol HCl}$

Step 4. Calculate the moles of ${\text{Al(OH)}}_{3}$

${\text{Moles of Al(OH)"_3 = 0.206 color(red)(cancel(color(black)("mol HCl"))) × ("1 mol Al(OH)"_3)/(3 color(red)(cancel(color(black)("mol HCl")))) = "0.0686 mol Al(OH)}}_{3}$

Step 5. Calculate the mass of ${\text{Al(OH)}}_{3}$

${\text{Mass of Al(OH)"_3 = 0.0686 color(red)(cancel(color(black)("mol Al(OH)"_3))) × ("78.00 g Al(OH)"_3)/( 1 color(red)(cancel(color(black)("mol Al(OH)"_3)))) = "5.35 g Al(OH)}}_{3}$

Step 6. Calculate the number of antacid tablets

$\text{Number of tablets" = 5.35 color(red)(cancel(color(black)("g Al(OH)"_3))) × "1 tablet"/(0.400 color(red)(cancel(color(black)("g Al(OH)"_3)))) = "13 tablets}$

1 (b)

Step 1. Write the balanced chemical equation for the reaction

$\text{Ca(OH"_2 + "2HCl" → "CaCl"_2 +2"H"_2"O}$

Step 2. Calculate the mass of the ${\text{Ca(OH)}}_{2}$ solution

$\text{Mass of solution" = 25 color(red)(cancel(color(black)("cm"^3color(white)(l) "solution"))) × "1.025 g solution"/(1 color(red)(cancel(color(black)("cm"^3 color(white)(l)"solution")))) = "25.6 g solution}$

Step 3. Calculate the mass of ${\text{Ca(OH)}}_{2}$

${\text{Mass of Ca(OH)"_2 = 25.6 color(red)(cancel(color(black)("g solution"))) × "1.85 g Ca(OH)"_2/(100 color(red)(cancel(color(black)("g solution")))) = "0.474 g Ca(OH)}}_{2}$

Step 4. Calculate the moles of ${\text{Ca(OH)}}_{2}$

${\text{Moles of Ca(OH)"_2 = 0.474 color(red)(cancel(color(black)("g Ca(OH)"_2))) × "1 mol Ca(OH)"_2/(74.09 color(red)(cancel(color(black)("g Ca(OH)"_2)))) = "0.006 40 mol Ca(OH)}}_{2}$

Step 5. Calculate the moles of $\text{HCl}$

$\text{Moles of HCl" = "0.006 40" color(red)(cancel(color(black)("mol Ca(OH)"_2))) × "2 mol HCl"/(1 color(red)(cancel(color(black)("mol Ca(OH)"_2)))) = "0.0128 mol HCl}$

Step 6. Calculate the volume of $\text{HCl}$

$\text{Volume of HCl" = 0.0128 color(red)(cancel(color(black)("mol HCl"))) × "1 L HCl"/(0.1 color(red)(cancel(color(black)("mol HCl")))) = "0.13 L HCl" = "130 mL HCl}$