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Chapter 6: Problem 59

What mass of solid aluminum hydroxide can be produced when 50.0 mL of \(0.200 M\) Al(NO_j) is added to 200.0 mL of \(0.100 M\) KOH?

Short Answer

Expert verified
Approximately 0.520 grams of solid aluminum hydroxide can be produced in this reaction.

Step by step solution

01

Determine the moles of each reactant

To determine the moles of each reactant, the volume (V) of the solutions and their molar concentration (c) should be used. The formula is: moles = V (in L) × c (in mol/L) For aluminum nitrate: moles of Al(NO₃)₃ = (50.0 mL × 0.200 mol/L) / 1000 moles of Al(NO₃)₃ = \( 0.0100 \) mol For potassium hydroxide: moles of KOH = (200.0 mL × 0.100 mol/L) / 1000 moles of KOH = \( 0.0200 \) mol
02

Identify the limiting reactant

In order to identify the limiting reactant, we can compare the mole ratio from the balanced chemical equation and the mole ratio of the reactants we are given. From the chemical equation, we know that: 1 mol Al(NO₃)₃ : 3 mol KOH Now, divide the moles of each reactant by their respective coefficients in the balanced chemical equation: For Al(NO₃)₃: \( \frac{0.0100 \text{ mol}}{1} = 0.0100 \) For KOH: \( \frac{0.0200 \text{ mol}}{3} = 0.00667 \) Since 0.00667 < 0.0100, KOH is the limiting reactant.
03

Calculate the moles of Al(OH)₃ produced

Now that we know KOH is the limiting reactant, we can calculate the moles of aluminum hydroxide formed. From the balanced chemical equation, the mole ratio is 1 mol Al(NO₃)₃ : 1 mol Al(OH)₃. moles of Al(OH)₃ = 0.00667 (from KOH) × 1 (ratio of Al(OH)₃ to KOH) moles of Al(OH)₃ = \( 0.00667 \) mol
04

Convert moles of Al(OH)₃ to mass

Finally, we will find the mass of aluminum hydroxide by converting the moles of Al(OH)₃ to grams using its molar mass. The molar mass of Al(OH)₃ is: 27.00 g/mol (Al) + 3 × (16.00 g/mol + 1.00 g/mol (OH component)) = 78.00 g/mol Now, we can convert the moles of Al(OH)₃ to grams: mass of Al(OH)₃ = \( 0.00667 \) mol × 78.00 g/mol mass of Al(OH)₃ ≈ 0.520 g So, approximately 0.520 grams of solid aluminum hydroxide can be produced in this reaction.

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Most popular questions from this chapter

A solution is prepared by dissolving \(10.8 \mathrm{g}\) ammonium sulfate in enough water to make \(100.0 \mathrm{mL}\) of stock solution. A \(10.00-\) mL sample of this stock solution is added to \(50.00 \mathrm{mL}\) of water. Calculate the concentration of ammonium ions and sulfate ions in the final solution.

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When organic compounds containing sulfur are burned, sulfur dioxide is produced. The amount of \(\mathrm{SO}_{2}\) formed can be determined by the reaction with hydrogen peroxide: $$ \mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{SO}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{SO}_{4}(a q) $$ The resulting sulfuric acid is then titrated with a standard NaOH solution. A 1.302 -g sample of coal is burned and the \(\mathrm{SO}_{2}\) is collected in a solution of hydrogen peroxide. It took \(28.44 \mathrm{mL}\) of a \(0.1000-M \mathrm{NaOH}\) solution to titrate the resulting sulfuric acid. Calculate the mass percent of sulfur in the coal sample. Sulfuric acid has two acidic hydrogens.

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