QuestionDetermine the number of grams of HCl needed to react with 2.99 g of $\mathrm{Al}(\mathrm{OH})_{3}$. $$\mathrm{Al}(\mathrm{OH})_{3}(\mathrm{~s})+3 \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{AlCl}_{3}(\mathrm{aq})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})$$ starting amount $\qquad$ $\qquad$ $\qquad$ $\qquad$

Studdy Solution
Calculate the mass of $\mathrm{HCl}$.
The molar mass of $\mathrm{HCl}$ is: - Atomic mass of H = 1.01 g/mol - Atomic mass of Cl = 35.45 g/mol
$\text{Molar mass of } \mathrm{HCl} = 1.01 + 35.45 = 36.46 \, \text{g/mol}$
$\text{Mass of } \mathrm{HCl} = 0.1149 \, \text{mol} \times 36.46 \, \text{g/mol} \approx 4.19 \, \text{g}$
The number of grams of $\mathrm{HCl}$ needed is:
$\boxed{4.19 \, \text{g}}$