Calculations Of Moles And Grams When Hydrochloric Acid Reacts With Aluminum Hydroxide
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Calculations Of Moles And Grams When Hydrochloric Acid Reacts With Aluminum Hydroxide

Hydrochloric Acid And Aluminum Hydroxide Reaction

Stomach acid is primarily hydrochloric acid (HCL). Several brands of antacid use Aluminum hydroxide (Al(OH)3 to react with stomach acid.
Given the following chemical reaction between hydrochloric acid and aluminum hydroxide:

Al(OH)3(s) + HCl(aq) ----> AlCl3(aq) + H2O(l).

(a) How many grams of HCl are needed to react with 0.500 grams of Al(OH)3?
(b) Calculate the number of grams AlCl3 and the number of grams of H2O formed when 0.500 g of Al(OH)3 reacts.
(c) Show that the solutions of (a) and (b) are consistent with the law of conservation of mass.

The strings: S7P6A66 (interaction - chemical).
S7P1A15 (containership - mass).

The math:
Although we are calculating , mass, grams and moles, the chemical reaction is the reason for the calculation. So Pj Problem of Interest is of type interaction (chemical).

Hydrocloric Acid Aluminum Hydroxide

(a) Molar mass of Al(OH)3 = 27 + 48 + 3 = 78
78 g of Al(OH)3 -----> 1 mole
So, 0.500 g of Al(OH)3 -----> 0.500/78 = 0.0064 mol
From chemical reaction:
1 mole of Al(OH)3 reacts with 3 moles of HCL
So, 0.0064 mol of Al(OH)3 will react with 3(0.0064) = 0.0192 mol of HCL
Molar mass of HCl = 1 + 35.5 = 36.5
So, weight of 0.0192 mol of HCL = 0.0192 x 36.5 = 0.701 g.

(b) 78 g of Al(OH)3 -----> 1 mole
So, 0.500 g of Al(OH)3 -----> 0.500/78 = 0.0064 mol
From chemical reaction:
1 mole of Al(OH)3 formed 1 mole of AlCl3
So, 0.0064 mol of Al(OH)3 will form 0.0064 mol of AlCl3
Molar mass of AlCl3 = 27 + 3(35.5) = 133.5
So, weight of 0.0064 mol of AlCl3 = 0.0192 x 133.5 = 0.855 g

1 mole of Al(OH)3 formed 3 moles of H2O
So, 0.0064 mol of Al(OH)3 will form 3(0.0064) mol of H2O = 0.0192 mol
Molar mass of H2O = 1 + 18 = 19
So, weight of 0.0192 mol of H2O = 18(0.0192) = 0.346 g

(c) Law of conservation of mass:
mass of reactants = mass of products
mass of reactants = 0.500 + 0.701 = 1.201 g
Mass of products = 0.855 + 0.346 = 1.201 g
So, calculations consistent with laws of conservation of mass.

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