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Chapter 4: Problem 60

A method of lowering the concentration of \(\mathrm{HCl}(\mathrm{aq})\) is to allow the solution to react with a small quantity of Mg. How many milligrams of Mg must be added to \(250.0 \mathrm{mL}\) of \(1.023 \mathrm{M} \mathrm{HCl}\) to reduce the solution concentration to exactly \(1.000 \mathrm{M} \mathrm{HCl} ?\) $$ \mathrm{Mg}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) \longrightarrow \mathrm{MgCl}_{2}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{g}) $$

Short Answer

Expert verified
69.9 mg of Mg must be added to 250.0 mL of 1.023 M HCl to reduce the solution concentration to exactly 1.000 M HCl.

Step by step solution

01

Calculate moles of HCl before reaction

First, use the formula for molarity which is \( M = \frac{n}{V} \), where \(M\) is molarity, \(n\) is the number of moles and \(V\) is the volume of the solution in liters. Calculation of moles of 1.023 M HCl in 250.0 ml = 1.023 mole/L \(\times\) 0.250 L = 0.25575 mole. So, there are initially 0.25575 moles of HCl in the solution.
02

Calculate moles of HCl after reaction

Then, calculate the amount of HCl needed to drop the concentration to 1.000 M in the 250 ml solution. Calculation: 1.000 M \(\times\) 0.250 L = 0.250 mole. So, after the reaction, there will be 0.250 moles of HCl left.
03

Calculate moles of HCl reacted

Subtract the moles of HCl after the reaction from the moles of HCl before the reaction to find the moles of HCl reacted. Calculation: 0.25575 mole - 0.250 mole = 0.00575 mole. So, 0.00575 moles of HCl reacted with the magnesium.
04

Calculate moles of Mg needed

Now, use the stoichiometry of the reaction to calculate the moles of Mg needed. From the reaction equation, 1 mole of Mg reacts with 2 mole of HCl. Therefore we get: 0.00575 mole of HCl \(\times\) \(\frac{1\,mole\,{Mg}}{2\,moles\,{HCl}} = 0.002875, which means 0.002875 moles of Mg are needed to lower the concentration of HCl to 1.000 M.
05

Convert moles of Mg to milligrams

Finally, convert the moles of Mg to milligrams using the molar mass of Mg which is 24.3050 g/mol. Calculation: 0.002875 moles of Mg \(\times\) \(\frac{24.3050\,g}{1\,mole}\) \(\times\) \(\frac{1000\,mg}{1\,g} = 69.8775 mg\). Rounding to three significant digits gives us 69.9 mg.

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Most popular questions from this chapter

To prepare a solution that is \(0.50 \mathrm{M} \mathrm{KCl}\) starting with \(100.0 \mathrm{mL}\) of \(0.40 \mathrm{M} \mathrm{KCl},\) you should \((\mathrm{a})\) add \(20.0 \mathrm{mL}\) of water; (b) add 0.075 g KCl; (c) add 0.10 mol KCl; (d) evaporate \(20.0 \mathrm{mL}\) of water.

Given two liters of \(0.496 \mathrm{M} \mathrm{KCl},\) describe how you would use this solution to prepare \(250.0 \mathrm{mL}\) of \(0.175 \mathrm{M} \mathrm{KCl} .\) Give sufficient details so that another student could follow your instructions.

Which of the following is a \(0.500 \mathrm{M} \mathrm{KCl}\) solution? (a) \(0.500 \mathrm{g} \mathrm{KCl} / \mathrm{mL}\) solution; (b) \(36.0 \mathrm{g} \mathrm{KCl} / \mathrm{L}\) solu- tion; (c) 7.46 mg KCl/mL solution; (d) 373 g KCl in 10.00 L solution

Solid calcium carbonate, \(\mathrm{CaCO}_{3}(\mathrm{s}),\) reacts with \(\mathrm{HCl}(\mathrm{aq})\) to form \(\mathrm{H}_{2} \mathrm{O}, \mathrm{CaCl}_{2}(\mathrm{aq}),\) and \(\mathrm{CO}_{2}(\mathrm{g}) .\) If a \(45.0 \mathrm{g}\) sample of \(\mathrm{CaCO}_{3}(\mathrm{s})\) is added to \(1.25 \mathrm{L}\) of \(\mathrm{HCl}(\mathrm{aq})\) that is \(25.7 \% \mathrm{HCl}\) by mass \((d=1.13 \mathrm{g} / \mathrm{mL})\) what will be the molarity of \(\mathrm{HCl}\) in the solution after the reaction is completed? Assume that the solution volume remains constant.

A side reaction in the manufacture of rayon from wood pulp is \(3 \mathrm{CS}_{2}+6 \mathrm{NaOH} \longrightarrow 2 \mathrm{Na}_{2} \mathrm{CS}_{3}+\mathrm{Na}_{2} \mathrm{CO}_{3}+3 \mathrm{H}_{2} \mathrm{O}\) How many grams of \(\mathrm{Na}_{2} \mathrm{CS}_{3}\) are produced in the reaction of \(92.5 \mathrm{mL}\) of liquid \(\mathrm{CS}_{2}(d=1.26 \mathrm{g} / \mathrm{mL})\) and 2.78 mol NaOH?
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