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

STEP 1

1. We are given the balanced chemical equation: $ \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}) \]
2. We have 2.99 grams of \(\mathrm{Al}(\mathrm{OH})_{3}\).
3. We need to calculate the mass of \(\mathrm{HCl}\) required to completely react with the given \(\mathrm{Al}(\mathrm{OH})_{3}\).

STEP 2

1. Calculate the molar mass of $\mathrm{Al}(\mathrm{OH})_{3}$ .
2. Calculate the moles of $\mathrm{Al}(\mathrm{OH})_{3}$ .
3. Use stoichiometry to find moles of $\mathrm{HCl}$ needed.
4. Calculate the mass of $\mathrm{HCl}$ .

STEP 3

Calculate the molar mass of $\mathrm{Al}(\mathrm{OH})_{3}$ .
- Atomic mass of Al = 26.98 g/mol - Atomic mass of O = 16.00 g/mol - Atomic mass of H = 1.01 g/mol
The molar mass of $\mathrm{Al}(\mathrm{OH})_{3}$ is calculated as follows:
$\text{Molar mass of } \mathrm{Al}(\mathrm{OH})_{3} = 26.98 + 3(16.00 + 1.01) = 26.98 + 3(17.01) = 26.98 + 51.03 = 78.01 \, \text{g/mol}$

STEP 4

Calculate the moles of $\mathrm{Al}(\mathrm{OH})_{3}$ .
$\text{Moles of } \mathrm{Al}(\mathrm{OH})_{3} = \frac{2.99 \, \text{g}}{78.01 \, \text{g/mol}} \approx 0.0383 \, \text{mol}$

STEP 5

Use stoichiometry to find moles of $\mathrm{HCl}$ needed.
From the balanced equation, 1 mole of $\mathrm{Al}(\mathrm{OH})_{3}$ reacts with 3 moles of $\mathrm{HCl}$ .
$\text{Moles of } \mathrm{HCl} = 0.0383 \, \text{mol} \times 3 = 0.1149 \, \text{mol}$

STEP 6

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:
$\boxed{4.19 \, \text{g}}$

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