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Chapter 4: Problem 89

A \(0.5895-\) g sample of impure magnesium hydroxide is dissolved in 100.0 \(\mathrm{mL}\) of 0.2050 \(\mathrm{M} \mathrm{HCl}\) solution. The excess acid then needs 19.85 \(\mathrm{mL}\) of 0.1020 \(\mathrm{M} \mathrm{NaOH}\) for neutralization. Calculate the percentage by mass of magnesium hydroxide in the sample, assuming that it is the only substance reacting with the HCl solution.

Short Answer

Expert verified
The percentage by mass of magnesium hydroxide in the sample is approximately 91.34%.

Step by step solution

01

Write down the balanced chemical equations

We have two reactions taking place here: 1. The reaction between magnesium hydroxide (Mg(OH)\(_2\)) and hydrochloric acid (HCl): \[ Mg(OH)_2 + 2HCl \rightarrow MgCl_2 + 2H_2O \] 2. The reaction between sodium hydroxide (NaOH) and the excess hydrochloric acid (HCl): \[ NaOH + HCl \rightarrow NaCl + H_2O \]
02

Calculate the moles of HCl initially present

We are given the volume (100.0 mL) and concentration (0.2050 M) of the HCl solution. To find the moles of HCl, we can use the formula: Moles of HCl = volume × concentration Moles of HCl = \( (100.0 \times 10^{-3} \, \text{L}) (0.2050 \, \text{M}) \) Moles of HCl = \( 0.02050 \, \text{mol} \)
03

Calculate the moles of HCl that reacted with Mg(OH)\(_2\)

We know that there are 19.85 mL of 0.1020 M NaOH needed to neutralize the excess HCl. We can find the moles of HCl reacted with NaOH as follows: Moles of HCl reacted with NaOH = volume × concentration Moles of HCl reacted with NaOH = \( (19.85 \times 10^{-3} \, \text{L}) (0.1020 \, \text{M}) \) Moles of HCl reacted with NaOH = \( 0.002023 \, \text{mol} \) Now, subtract this amount of HCl from the initial moles of HCl to find the moles of HCl that reacted with Mg(OH)\(_2\): Moles of HCl reacted with Mg(OH)\(_2\) = Initial moles of HCl - Moles of HCl reacted with NaOH Moles of HCl reacted with Mg(OH)\(_2\) = 0.02050 - 0.002023 Moles of HCl reacted with Mg(OH)\(_2\) = \( 0.01848 \, \text{mol} \)
04

Calculate the moles of Mg(OH)\(_2\) in the sample

Using the stoichiometry of the balanced equation of reaction between Mg(OH)\(_2\) and HCl (1:2 ratio), we can find the moles of Mg(OH)\(_2\): Moles of Mg(OH)\(_2\) = Moles of HCl reacted with Mg(OH)\(_2\) / 2 Moles of Mg(OH)\(_2\) = 0.01848 / 2 Moles of Mg(OH)\(_2\) = \( 0.00924 \, \text{mol} \)
05

Calculate the mass of Mg(OH)\(_2\) in the sample

To find the mass of Mg(OH)\(_2\), we multiply its moles by its molar mass: Mass of Mg(OH)\(_2\) = moles × molar mass Molar mass of Mg(OH)\(_2\) = 24.31 (Mg) + 2(16.00 + 1.01) (2OH) = 58.33 g/mol Mass of Mg(OH)\(_2\) = \( (0.00924 \, \text{mol}) (58.33 \, \text{g/mol}) \) Mass of Mg(OH)\(_2\) = \( 0.5385 \, \text{g} \)
06

Calculate the percentage by mass of Mg(OH)\(_2\) in the sample

Finally, we can find the percentage by mass of Mg(OH)\(_2\) in the sample using the formula: Percentage by mass of Mg(OH)\(_2\) = (Mass of Mg(OH)\(_2\) / Total mass of the sample) × 100 Percentage by mass of Mg(OH)\(_2\) = \( (0.5385 / 0.5895) \times 100 \) Percentage by mass of Mg(OH)\(_2\) = \( 91.34 \% \) Thus, the percentage by mass of magnesium hydroxide in the sample is approximately 91.34%.

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