Question

Aluminum hydroxide $\left(\mathrm{Al}(\mathrm{OH})_{3}\right)$ is often present in antacids to neutralize stomach acid (HCl). The reaction occurs as follows: $\mathrm{Al}(\mathrm{OH})_{3}(\mathrm{s})+3 \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{AlCl}_{3}(\mathrm{aq})+3 \mathrm{H}_{2} \mathrm{O}(1) .$ If 14.0 $\mathrm{g}$ of $\mathrm{Al}(\mathrm{OH})_{3}$ is present in an antacid tablet, determine the theoretical yield of AlCl_ produced when the tablet reacts with HCl.

          Aluminum hydroxide $\left(\mathrm{Al}(\mathrm{OH})_{3}\right)$ is often present in antacids to
neutralize stomach acid (HCl). The reaction occurs as follows:
$\mathrm{Al}(\mathrm{OH})_{3}(\mathrm{s})+3 \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{AlCl}_{3}(\mathrm{aq})+3 \mathrm{H}_{2} \mathrm{O}(1) .$ If 14.0 $\mathrm{g}$ of $\mathrm{Al}(\mathrm{OH})_{3}$ is present in an antacid tablet, determine the theoretical yield of
AlCl_ produced when the tablet reacts with HCl.

Added by Edward C.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Adi S

03/14/2023

Step 1

The molar mass of $\mathrm{Al}(\mathrm{OH})_{3}$ is: $1 \times \mathrm{Al} + 3 \times (\mathrm{O} + \mathrm{H}) = 1 \times 26.98 + 3 \times (16.00 + 1.01) = 26.98 + 3 \times 17.01 = 26.98 + 51.03 = 78.01 \ \mathrm{g/mol}$ The molar mass of $\mathrm{AlCl}_{3}$ Show more…

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Aluminum hydroxide $\left(\mathrm{Al}(\mathrm{OH})_{3}\right)$ is often present in antacids to neutralize stomach acid (HCl). The reaction occurs as follows: $\mathrm{Al}(\mathrm{OH})_{3}(\mathrm{s})+3 \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{AlCl}_{3}(\mathrm{aq})+3 \mathrm{H}_{2} \mathrm{O}(1) .$ If 14.0 $\mathrm{g}$ of $\mathrm{Al}(\mathrm{OH})_{3}$ is present in an antacid tablet, determine the theoretical yield of AlCl_ produced when the tablet reacts with HCl.