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When hydrochloric acid reacts with magnesium metal, hydrogen gas and aqueous magnesium chloride are produced. What volume of \(5.0 \mathrm{M} \mathrm{HCl}\) is required to react completely with \(3.00 \mathrm{~g}\) of magnesium?

Short Answer

Expert verified
To react completely with 3.00 g of magnesium, 49.2 mL of 5.0 M HCl solution is required.

Step by step solution

01

Write the balanced chemical equation for the reaction

We are given that hydrochloric acid (HCl) reacts with magnesium (Mg) to produce hydrogen gas (H₂) and aqueous magnesium chloride (MgCl₂). The balanced equation for this reaction is: \[ Mg(s) + 2HCl(aq) \rightarrow H_{2}(g) + MgCl_{2}(aq) \]
02

Convert mass of magnesium to moles

We're given 3.00 g of magnesium. To determine the moles, we need to use its molar mass, which is 24.31 g/mol. \[ moles \: of \: Mg = \frac{3.00 \: g}{24.31 \: g/mol} \] \[ moles \: of \: Mg = 0.123 \: mol \]
03

Use the stoichiometry to find the moles of HCl required

From the balanced chemical equation, we can see that 1 mole of magnesium (Mg) reacts with 2 moles of hydrochloric acid (HCl). Using the moles of Mg we just calculated, we can determine the moles of HCl required. \[ moles \: of \: HCl = 0.123 \: mol \times \frac{2 \: mol \: HCl}{1 \: mol \: Mg} \] \[ moles \: of \: HCl = 0.246 \: mol \]
04

Calculate the volume of 5.0 M HCl solution required

We know the molarity of the HCl solution (5.0 M) and the moles of HCl required for the reaction (0.246 mol). Using the formula for molarity, we can determine the volume of the 5.0 M HCl solution required for this reaction. The formula for molarity is: \[ M = \frac{moles \: of \: solute}{volume \: of \: solution (L)} \] Rearranging the formula to find the volume, we get: \[ volume \: of \: solution (L) = \frac{moles \: of \: solute}{M} \] Now we can plug in the values: \[ volume \: of \: solution (L) = \frac{0.246 \: mol}{5.0 \: M} \] \[ volume \: of \: solution (L) = 0.0492 \: L \] Since 1 L = 1000 mL, we can convert the volume to milliliters: \[ volume \: of \: solution (mL) = 0.0492 \: L \times 1000 \: mL/L \] \[ volume \: of \: solution (mL) = 49.2 \: mL \] So, 49.2 mL of 5.0 M HCl solution is required to react completely with 3.00 g of magnesium.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Reaction Equations
Chemical reaction equations are symbolic representations of chemical reactions. They use chemical formulas to denote the reactants and products, and coefficients to reflect the quantities that balance the reaction in accordance with the law of conservation of mass. The equation tells us not only which substances react and what products they form, but also their relative amounts.

For example, in the reaction of hydrochloric acid with magnesium, the equation is written as \[Mg(s) + 2HCl(aq) \rightarrow H_{2}(g) + MgCl_{2}(aq)\]
The reactants, magnesium (Mg) and hydrochloric acid (HCl), are on the left, while the products, hydrogen gas (H₂) and magnesium chloride (MgCl₂), are on the right. The '(s)', '(aq)', and '(g)' symbols indicate the physical states of the substances: solid, aqueous (dissolved in water), and gas, respectively. The coefficient '2' before HCl indicates that two moles of HCl react with one mole of Mg. Understanding the equation is vital for correctly performing the stoichiometric calculations detailed in subsequent steps.
Molar Mass Calculations
Molar mass is an essential concept when working with chemical substances because it relates the mass of a substance to its number of particles, usually moles. Every element has a characteristic molar mass which can be found on the periodic table, and for compounds, it's the sum of the molar masses of its constituent elements.

In the given exercise, we calculated the molar mass of magnesium (\(24.31 \text{g/mol}\)). To find out how many moles are contained in a given mass of magnesium (in this case, \(3.00 \text{g}\)), we divide the mass by the molar mass, as shown below:\[ moles \: of \: Mg = \frac{3.00 \: g}{24.31 \: g/mol} \]\[ moles \: of \: Mg = 0.123 \: mol \]This conversion is crucial in stoichiometry because chemical reactions occur in proportion to the moles of reactants and products, not just their mass.
Molarity and Solution Volume
Molarity (M) is a measure of the concentration of a solute in a solution. It is defined as the number of moles of solute per liter of solution. The molarity can help us find out how much of a substance is present in a certain volume of liquid or, conversely, how much volume is needed for a certain amount of the substance.

To calculate the volume of solution required for a known mole of solute in a reaction, we reorganize the formula for molarity, which looks like this:\[ M = \frac{moles \: of \: solute}{volume \: of \: solution (L)} \] Rearranging the formula to find the volume, we get:\[ volume \: of \: solution (L) = \frac{moles \: of \: solute}{M} \] For instance, to react with \(0.246 \: mol\) of HCl of a 5.0 M HCl solution, we need to divide the moles of HCl by the concentration of the solution:\[ volume \: of \: solution (L) = \frac{0.246 \: mol}{5.0 \: M} \]\[ volume \: of \: solution (L) = 0.0492 \: L \] This relationship between molarity, volume, and moles is central to preparing solutions and executing reactions in chemistry.
Reactant to Product Conversion
In stoichiometry, reactant to product conversion is the process by which we use the balanced chemical equation to determine the amount of product formed from a given quantity of reactant, or vice versa. It is based on the stoichiometric coefficients of the reactants and products which represent their mole ratios in the reaction.

Through the stoichiometry of the balanced reaction, we can find out that 1 mole of magnesium requires 2 moles of hydrochloric acid to fully react as shown in the following step:\[ moles \: of \: HCl = 0.123 \: mol \times \frac{2 \: mol \: HCl}{1 \: mol \: Mg} \]\[ moles \: of \: HCl = 0.246 \: mol \]This conversion ensures that the reaction proceeds with the correct proportions, assuring that all of the reactant is used up, provided we have the necessary amounts of the other reactants. This core concept is integral for predicting how much product we can expect from a reaction and determining all reactant amounts needed.

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Most popular questions from this chapter

Calculate the sodium ion concentration when \(70.0 \mathrm{~mL}\) of 3.0 \(M\) sodium carbonate is added to \(30.0 \mathrm{~mL}\) of \(1.0 \mathrm{M}\) sodium bicarbonate.

Describe how you would prepare \(2.00 \mathrm{~L}\) of each of the following solutions. a. \(0.250 \mathrm{M} \mathrm{NaOH}\) from solid \(\mathrm{NaOH}\) b. \(0.250 M \mathrm{NaOH}\) from \(1.00 M \mathrm{NaOH}\) stock solution c. \(0.100 \mathrm{M} \mathrm{K}_{2} \mathrm{CrO}_{4}\) from solid \(\mathrm{K}_{2} \mathrm{CrO}_{4}\) d. \(0.100 M \mathrm{~K}_{2} \mathrm{CrO}_{4}\) from \(1.75 \mathrm{M} \mathrm{K}_{2} \mathrm{CrO}_{4}\) stock solution

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How would you prepare \(1.00 \mathrm{~L}\) of a \(0.50-M\) solution of each of the following? a. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) from "concentrated" ( \(18 \mathrm{M}\) ) sulfuric acid b. HCl from "concentrated" (12 \(M\) ) reagent c. \(\mathrm{NiCl}_{2}\) from the salt \(\mathrm{NiCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) d. HNO \(_{3}\) from "concentrated" ( \(16 M\) ) reagent e. Sodium carbonate from the pure solid

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