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 unchanged.

Short Answer

Expert verified
The concentration of the acid solution after all the magnesium has reacted is approximately \(1.26 \mathrm{M}\).

Step by step solution

01

Find the Moles of Magnesium

First, the moles of magnesium (Mg) need to be calculated using the conversion factor from the periodic table (24.31g/mol). Use the formula: \( \text{moles} = \frac{\text{mass}}{\text{molar mass}}\). With mass: 4.47g and molar mass: 24.31g/mol, the calculation for moles of magnesium is: \( \text{moles Mg} = \frac{4.47g}{24.31g/mol} \approx 0.184 \text{ moles}\).
02

Calculate the Moles of Hydrochloric Acid That Reacted

From the balanced reaction equation: \(2 \mathrm{HCl} + \mathrm{Mg} \rightarrow \mathrm{H_2} + \mathrm{MgCl_2}\), it is understood that 2 moles of HCl react with 1 mole of Mg. Therefore, \(0.184 \text{ moles}\) of Mg react with \(2 \times 0.184 \text{ moles of HCl} = 0.368 \text{moles of HCl}\).
03

Calculate the Initial Moles of Hydrochloric Acid in Solution

The initial moles of acid can be calculated using the formula: \( \text{Moles} = \text{Molarity} \times \text{Volume}\). Since the volume is given in mL and for this calculation we need the volume in liters, convert 500mL to 0.5L. So: \( \text{Initial moles of HCl} = 2.00M \times 0.5L = 1.00 \text{ moles}\).
04

Calculate the Final Concentration of the Solution

The final moles of HCl left after the reaction with magnesium can be calculated by subtracting moles of HCl that reacted from the initial moles of HCl. So: \( \text{Final moles of HCl} = \text{Initial moles of HCl} - \text{moles of HCl reacted} = 1.00 \text{ mol} - 0.368 \text{ mol} = 0.632 \text{ mol}\). The final concentration (Molarity) of HCl is given by dividing the remaining moles by the volume of the solution in Liters: \( \text{Final Molarity} = \frac{\text{Final moles of HCl}}{\text{Volume}} = \frac{0.632 \text{ mol}}{0.5L} \approx 1.26M\).

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Most popular questions from this chapter

Magnesium is a valuable, lightweight metal. It is used as a structural metal and in alloys, in batteries, and in chemical synthesis. Although magnesium is plentiful in Earth's crust, it is cheaper to "mine" the metal from seawater. Magnesium forms the second most abundant cation in the sea (after sodium); there are about \(1.3 \mathrm{~g}\) of magnesium in \(1 \mathrm{~kg}\) of seawater. The method of obtaining magnesium from seawater employs all three types of reactions discussed in this chapter: precipitation, acid-base, and redox reactions. In the first stage in the recovery of magnesium, limestone \(\left(\mathrm{CaCO}_{3}\right)\) is heated at high temperatures to produce quicklime, or calcium oxide \((\mathrm{CaO})\) : $$ \mathrm{CaCO}_{3}(s) \longrightarrow \mathrm{CaO}(s)+\mathrm{CO}_{2}(g) $$ When calcium oxide is treated with seawater, it forms calcium hydroxide \(\left[\mathrm{Ca}(\mathrm{OH})_{2}\right]\), which is slightly soluble and ionizes to give \(\mathrm{Ca}^{2+}\) and \(\mathrm{OH}^{-}\) ions: $$ \mathrm{CaO}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}^{2+}(a q)+2 \mathrm{OH}^{-}(a q) $$ The surplus hydroxide ions cause the much less soluble magnesium hydroxide to precipitate: $$ \mathrm{Mg}^{2+}(a q)+2 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{Mg}(\mathrm{OH})_{2}(s) $$ The solid magnesium hydroxide is filtered and reacted with hydrochloric acid to form magnesium chloride \(\left(\mathrm{MgCl}_{2}\right)\) \(\mathrm{Mg}(\mathrm{OH})_{2}(s)+2 \mathrm{HCl}(a q) \longrightarrow\) $$ \mathrm{MgCl}_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ After the water is evaporated, the solid magnesium chloride is melted in a steel cell. The molten magnesium chloride contains both \(\mathrm{Mg}^{2+}\) and \(\mathrm{Cl}^{-}\) ions. In a process called electrolysis, an electric current is passed through the cell to reduce the \(\mathrm{Mg}^{2+}\) ions and oxidize the \(\mathrm{Cl}^{-}\) ions. The halfreactions are $$ \begin{aligned} \mathrm{Mg}^{2+}+2 e^{-} \longrightarrow \mathrm{Mg} \\ 2 \mathrm{Cl}^{-} \longrightarrow \mathrm{Cl}_{2}+2 e^{-} \end{aligned} $$ The overall reaction is $$ \mathrm{MgCl}_{2}(l) \longrightarrow \mathrm{Mg}(s)+\mathrm{Cl}_{2}(g) $$ This is how magnesium metal is produced. The chlorine gas generated can be converted to hydrochloric acid and recycled through the process. (a) Identify the precipitation, acid-base, and redox processes. (b) Instead of calcium oxide, why don't we simply add sodium hydroxide to precipitate magnesium hydroxide? (c) Sometimes a mineral called dolomite (a combination of \(\mathrm{CaCO}_{3}\) and \(\mathrm{MgCO}_{3}\) ) is substituted for limestone \(\left(\mathrm{CaCO}_{3}\right)\) to bring about the precipitation of magnesium hydroxide. What is the advantage of using dolomite? (d) What are the advantages of mining magnesium from the ocean rather than from Earth's crust?

Give an example of a monoprotic acid, a diprotic acid, and a triprotic acid.

Balance the following equations and write the corresponding ionic and net ionic equations (if appropriate): (a) \(\mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{KOH}(a q) \longrightarrow\) (b) \(\mathrm{H}_{2} \mathrm{CO}_{3}(a q)+\mathrm{NaOH}(a q) \longrightarrow\) (c) \(\mathrm{HNO}_{3}(a q)+\mathrm{Ba}(\mathrm{OH})_{2}(a q) \longrightarrow\)

Identify each of the following species as a Bronsted acid, base, or both: (a) HI, (b) \(\mathrm{CH}_{3} \mathrm{COO}^{-}\), (c) \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) (d) \(\mathrm{HSO}_{4}^{-}\)

(a) Without referring to Figure 4.10 , give the oxidation numbers of the alkali and alkaline earth metals in their compounds. (b) Give the highest oxidation numbers that the Groups \(3 \mathrm{~A}-7 \mathrm{~A}\) elements can have.

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