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Chapter 4: Problem 106

A \(5.00 \times 10^{2} \mathrm{~mL}\) sample of \(2.00 \mathrm{M} \mathrm{HCl}\) solution is treated with \(4.47 \mathrm{~g}\) of magnesium. Calculate the concentration of the acid solution after all the metal has reacted. Assume that the volume remains un- changed.

Short Answer

Expert verified
The concentration of the Hydrochloric Acid solution after all the Magnesium has reacted is \(1.2648 \mathrm{M}\)

Step by step solution

01

Calculate the moles of Magnesium

Moles of a substance can be calculated by dividing the mass given by the molar mass. So, for Magnesium, it would be \(\frac{4.47 \mathrm{g}}{24.31 \mathrm{g/mol}} = 0.1838 \mathrm{mol}\)
02

Determine the moles of Hydrochloric Acid

The molarity of a solution is defined as the number of moles of solute per litre of solution. Since we are given that the initial concentration of the acid solution is 2.00 M and volume is \(500.00 \mathrm{mL} = 0.500 \mathrm{L}\), multiplying the two would give the initial moles of Hydrochloric Acid. So, \(2.00 \mathrm{M} \times 0.500 \mathrm{L} = 1.00 \mathrm{mol}\)
03

Determine the remaining moles of Hydrochloric Acid

Since 1 mol of Magnesium reacts with 2 mol of Hydrochloric acid, and the number of moles of Magnesium is 0.1838 mol, multiply this number by 2 to get the number of moles of acid that will react with Magnesium, which gives \(0.1838 \mathrm{mol} \times 2 = 0.3676 \mathrm{mol}\). Subtracting this number from the initial moles of Hydrochloric Acid will give the remaining moles. Thus, \(1.00 \mathrm{mol} - 0.3676 \mathrm{mol} = 0.6324 \mathrm{mol}\)
04

Calculate the final concentration of Hydrochloric Acid

The final concentration can be calculated by dividing the remaining moles of Hydrochloric Acid by the volume of the solution in litres. So, \( \frac{0.6324 \mathrm{mol}}{0.500 \mathrm{L}} = 1.2648 \mathrm{M}\)

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Most popular questions from this chapter

Give a chemical explanation for each of the following: (a) When calcium metal is added to a sulfuric acid solution, hydrogen gas is generated. After a few minutes, the reaction slows down and eventually stops even though none of the reactants is used up. (b) In the activity series, aluminum is above hydrogen, yet the metal appears to be unreactive toward steam and hydrochloric acid. (c) Sodium and potassium lie above copper in the activity series. In your explanation, discuss why \(\mathrm{Cu}^{2+}\) ions in a \(\mathrm{CuSO}_{4}\) solution are not converted to metallic copper upon the addition of these metals. (d) A metal M reacts slowly with steam. There is no visible change when it is placed in a pale green iron(II) sulfate solution. Where should we place \(\mathrm{M}\) in the activity series? (e) Before aluminum metal was obtained by electrolysis, it was produced by reducing its chloride \(\left(\mathrm{AlCl}_{3}\right)\) with an active metal. What metals would you use to produce aluminum in that way?

Chemical tests of four metals \(\mathrm{A}, \mathrm{B}, \mathrm{C},\) and \(\mathrm{D}\) show the following results. (a) Only \(\mathrm{B}\) and \(\mathrm{C}\) react with \(0.5 \mathrm{M} \mathrm{HCl}\) to give \(\mathrm{H}_{2}\) gas. (b) When \(\mathrm{B}\) is added to a solution containing the ions of the other metals, metallic \(\mathrm{A}, \mathrm{C},\) and \(\mathrm{D}\) are formed. (c) A reacts with \(6 M \mathrm{HNO}_{3}\) but \(\mathrm{D}\) does not. Arrange the metals in the increasing order as reducing agents. Suggest four metals that fit these descriptions.

Describe how you would prepare the following compounds: (a) \(\mathrm{Mg}(\mathrm{OH})_{2},\) (b) \(\mathrm{AgI}\), (c) \(\mathrm{Ba}_{3}\left(\mathrm{PO}_{4}\right)_{2}\).

A student carried out two titrations using a \(\mathrm{NaOH}\) solution of unknown concentration in the buret. In one titration she weighed out \(0.2458 \mathrm{~g}\) of KHP (see Section 4.7 ) and transferred it to an Erlenmeyer flask. She then added \(20.00 \mathrm{~mL}\) of distilled water to dissolve the acid. In the other titration she weighed out \(0.2507 \mathrm{~g}\) of KHP but added \(40.00 \mathrm{~mL}\) of distilled water to dissolve the acid. Assuming no experimental error, would she obtain the same result for the concentration of the \(\mathrm{NaOH}\) solution?

A \(35.2-\mathrm{mL}, 1.66 M \mathrm{KMnO}_{4}\) solution is mixed with \(16.7 \mathrm{~mL}\) of \(0.892 \mathrm{M} \mathrm{KMnO}_{4}\) solution. Calculate the concentration of the final solution.
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