What is the percent by mass of the magnesium hydroxide?

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What is the percent by mass of the magnesium hydroxide?

A 0.5895 g sample of impure magnesium hydroxide is dissolved in 100.0 mL of 0.2050 M HCl solution. The excess acid needs 19.85 mL of 0.1020 M NaOH for neutralization.

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Answer 1 · 2016-05-15T12:31:29

The sample is 91.42% magnesium hydroxide.

Explanation:

Let us first work out how many moles of acid were neutralized by the sample.

Beginning:

$0.1 l \times \frac{0.2050 mol HCl}{l} = 0.02050$ $0.1 l\times{0.2050" mol HCl"}/l=0.02050$ mol $HCl$ $"HCl"$

After the sample was added:

We are given the amount of $NaOH$ $"NaOH"$ that would neutralize the remaining $HCl$ $"HCl"$ :

$0.01985 l \times \frac{0.1020 mol NaOH}{l} = 0.0020247$ $0.01985 l\times{0.1020 "mol NaOH"}/l=0.0020247$ mol $NaOH$ $"NaOH"$

Now each me of the added $NaOH$ $"NaOH"$ neutralized one mole of leftover $HCl$ $"HCl"$ from the balanced equation

$NaOH + HCl \to NaCl + H_{2} O$ $"NaOH"+"HCl"->"NaCl"+"H"_2"O"$

so we have:

$0.0020247$ $0.0020247$ mol $NaOH \times \frac{1 mol HCl}{1 mol NaOH} = 0.0020247$ $"NaOH"\times{1" mol HCl"}/{1" mol NaOH"}=0.0020247$ mol $HCl$ $"HCl"$ left over

So we started with 0.02050 mol $HCl$ $"HCl"$ and we were left with 0.0020247 mol $HCl$ $"HCl"$ to after the sample was added. The difference, rounded to the smaller number of decimal places, is:

$0.02050 - 0.00202 = 0.01848 mol HCl$ $0.02050-0.00202=0.01848 "mol HCl"$ that the sample neutralized.

Now we can work out how mych magnesium hydroxide must have neutealuzed that acid. Start with the balanced equation:

${Mg(OH)}_{2} + 2 HCl \to {MgCl}_{2} + 2 H_{2} O$ $"Mg(OH)"_2+2 "HCl"->"MgCl"_2+2 "H"_2"O"$

So then:

$0.01848 mol HCl \times \frac{1 {mol Mg(OH)}_{2}}{2 mol HCl} = 0.009240 {mol Mg(OH)}_{2}$ $0.01848" mol HCl"\times{1" mol Mg(OH)"_2}/{2" mol HCl"}=0.009240 "mol Mg(OH)"_2$ in the sample

We weighed the sample in grams so calculate the grams of magnesium hydroxide using the molecular or formula weight $= 58.3197 g/mol$ $=58.3197" g/mol"$ . So

$0.009240 mol \times 58.3197 g/mol = 0.5389 g$ $0.009240 "mol"\times58.3197" g/mol"=0.5389 g$ .

This is how much magnesium hydroxide is in the $0.5895 g$ $0.5895 "g"$ 'sample. Then

$0.5389 \frac{g}{0.5895} g = 0.9142 = 91.42 %$ $0.5389 "g"/0.5895"g"=0.9142=91.42%$

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What is the percent by mass of the magnesium hydroxide?

A 0.5895 g sample of impure magnesium hydroxide is dissolved in 100.0 mL of 0.2050 M HCl solution. The excess acid needs 19.85 mL of 0.1020 M NaOH for neutralization.

1 Answer

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