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How many grams of zinc are reguired to reduce Fe3+ when \(25.0 \mathrm{~mL}\) of \(1.2 \mathrm{M} \mathrm{FeCl}\) are reacted? $$ \mathrm{Zn}+\mathrm{Fe}^{3+} \rightarrow \mathrm{Zn}^{2+}+\mathrm{Fe}

Short Answer

Expert verified
1.96 grams of Zn are required.

Step by step solution

01

Calculate moles of Fe³⁺

First, determine the number of moles of Fe³⁺ in the solution. Use the formula \text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}}. Given that the molarity (\text{M}) is 1.2M and the volume is 25.0 mL (0.025 L), the moles of Fe³⁺ can be calculated as follows: \[ \text{moles of Fe³⁺} = 1.2 \frac{\text{moles}}{L} \times 0.025 L = 0.03 \text{ moles} \text{ of Fe³⁺} \]
02

Determine the moles of Zn required

Using the balanced chemical equation: \[ \text{Zn} + \text{Fe}^{3+} \rightarrow \text{Zn}^{2+} + \text{Fe} \] it can be seen that 1 mole of Zn is required to reduce 1 mole of Fe³⁺. Therefore, 0.03 moles of Fe³⁺ will require 0.03 moles of Zn.
03

Convert moles of Zn to grams

Finally, convert the moles of Zn needed to grams using the molar mass of zinc. The molar mass of Zn is 65.38 g/mol. The mass of Zn required is given by: \[ \text{mass of Zn} = 0.03 \text{ moles} \times 65.38 \frac{\text{g}}{\text{mol}} = 1.9614 \text{ g} \] Round this to 1.96 grams.

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Key Concepts

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

chemical reactions
Chemical reactions involve transforming reactants into products. The chemical equation provided, \(\text{Zn} + \text{Fe}^{3+} \rightarrow \text{Zn}^{2+} + \text{Fe}\), illustrates a simple displacement reaction. Zinc (Zn) reduces iron ions (Fe³⁺) to form zinc ions (Zn²⁺) and solid iron (Fe). When balancing chemical equations, it's vital to ensure that the number of atoms for each element is the same on both sides of the equation. Here, we see one atom of zinc and one atom of iron change their oxidation states. This balanced equation helps in understanding the stoichiometric relationships needed for calculations.
molarity
Molarity (M) is a measure of the concentration of a solution. It's defined as the number of moles of solute (the substance dissolved) per liter of solution. The formula is: \(\text{Molarity} = \frac{\text{moles of solute}}{\text{volume of solution in liters}}\). For example, in the given problem, the solution's molarity is 1.2 M and its volume is 25.0 mL (0.025 L). Using these values: \[ \text{moles of Fe}^{3+} = 1.2 \frac{\text{moles}}{\text{L}} \times 0.025 \text{ L} = 0.03 \text{ moles} \] Understanding molarity is crucial for determining the amount of solute in a given volume of solution.
mole-to-mole conversions
Mole-to-mole conversions rely on the coefficients in a balanced chemical equation. These coefficients indicate the ratio in which reactants combine and products form. In the reaction \(\text{Zn} + \text{Fe}^{3+} \rightarrow \text{Zn}^{2+} + \text{Fe}\), the coefficients (implied to be 1 for all reactants and products) show that one mole of zinc reacts with one mole of Fe³⁺. Thus, if you have 0.03 moles of Fe³⁺, it will require 0.03 moles of zinc to react completely with it, applying the 1:1 ratio.
mass calculations
To find the mass of a substance, you multiply its moles by its molar mass (the mass of one mole of that substance). The molar mass of zinc is 65.38 g/mol. Given that we need 0.03 moles of zinc, the calculation is: \[ \text{mass of Zn} = 0.03 \text{ moles} \times 65.38 \frac{\text{g}}{\text{mol}} = 1.9614 \text{ g} \] This can be rounded to 1.96 grams. This step ensures we translate moles back into a measurable mass, finalizing our stoichiometry calculation.

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Most popular questions from this chapter

Determine whether each of the following half-reactions represents an oxidation or a reduction. Supply the correct number of electrons to the appropriate side to balance the cquation. (a) \(\mathrm{Cu}^{2+} \longrightarrow \mathrm{Cu}^{1+}\) (b) \(\mathrm{F}_{2} \rightarrow 2 \mathrm{~F}^{-}\) (c) \(2 \mathrm{IO}_{4}^{-}+16 \mathrm{H}^{+} \longrightarrow \mathrm{I}_{2}+8 \mathrm{H}_{2} \mathrm{O}\) (d) \(\mathrm{Mn} \longrightarrow \mathrm{Mn}^{2+}\)

Determine whether the following oxidation-reduction reactions ate balanced correctly, If they are not, provide the correct balanced reaction. (a) unbalanced: \(\mathrm{MnO}_{2}(s)+\mathrm{Al}(s) \longrightarrow \mathrm{Mn}(s)+\mathrm{Al} \mathrm{O}_{3}(s)\) balanced: \(\mathrm{MnO}_{2}(s)+2 A 1(s) \longrightarrow \mathrm{Mn}(s)+\mathrm{Al}_{2} \mathrm{O}_{3}(s)\) (b) unbalanced: \(\mathrm{Cu}(s)+\mathrm{Ag}^{4}(a q) \longrightarrow \mathrm{Cu}^{24}(a q)+A \mathrm{~g}(s)\) balanced: \(\mathrm{Cu}(s)+2 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Cu}^{2+}(a q)+2 \mathrm{Ag}(s)\) (c) unbalanced: $$ \mathrm{Br}^{-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{Br}_{2}(d)+\mathrm{Mn}^{2+}(a q) $$ (icidic solution) balanced: $$ \begin{aligned} 16 \mathrm{H}^{+}(a q)+10 \mathrm{Br}^{-}(a q)+& 2 \mathrm{MnO}_{4}^{-}(a q) \longrightarrow \\ & 5 \mathrm{Br}(l)+2 \mathrm{Mn}^{2+}(a q)+8 \mathrm{H}_{2} \mathrm{O}(l) \end{aligned} $$ (d) unhalanced: $$ \begin{aligned} &\mathrm{MnO}_{4}(a q)+\mathrm{S}^{2-}(a q) \rightarrow \mathrm{MnS}(s)+\mathrm{S}(s) \quad \text { (basic solution) } \\ &\text { balanced: } \\ &8 \mathrm{H}^{-1}(a q)+\mathrm{MnO}_{4}^{-}(a q)+\mathrm{S}^{2-}(a q) \rightarrow \\ &\qquad \mathrm{S}(s)+\mathrm{MnS}(s)+4 \mathrm{H}_{2} \mathrm{O}(l) \end{aligned} $$

Balance each of the following equations using the change-inoxidation-number method: (a) \(\mathrm{Cu}+\mathrm{O}_{2} \rightarrow \mathrm{CuO}\) (b) \(\mathrm{KClO}_{2} \rightarrow \mathrm{KCl}+\mathrm{O}_{2}\) (c) \(\mathrm{Ca}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{Ca}(\mathrm{OH})_{2}+\mathrm{H}_{2}\) (d) \(\mathrm{PbS}+\mathrm{H}_{2} \mathrm{O}_{2} \rightarrow \mathrm{PbSO}_{4}+\mathrm{H}_{2} \mathrm{O}\) (e) \(\mathrm{CH}_{4}+\mathrm{NO}_{2} \rightarrow \mathrm{N}_{2}+\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}\)

(a) \(\mathrm{Al}\) is oxidized and loses 3 electrons. \(\mathrm{H}\) is reduced and gains 1 electron. \(\mathrm{Al}\) is the reducing agent and \(\mathrm{HNO}_{3}\) is the oxidizing agent. (b) Ni is oxidized and loses 2 electrons. Cu is reduced and gains 2 electrons. \(\mathrm{Ni}\) is the reducing agent and \(\mathrm{CuSO}_{4}\) is the axidioing agent. (c) C is oxidized and loses 1 electron. \(O\) is reduced and gains 2 electrons. \(\mathrm{C}_{2} \mathrm{H}_{6}\) is the reducing agent and \(\mathrm{O}_{2}\) is the oxidizing agent.

Alcohols can be oxidized very easily by potassium dichromate. This fact formed the basis for many of the early breathalyzers used by law eaforcement officers to determine the approximate blood alcohol levels of suspected drunk drivers. The unbalanced equation for the reaction is $$ \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}+\mathrm{Cr}_{2} \mathrm{O}^{2} \rightarrow \mathrm{CH}_{3} \mathrm{COOH}+\mathrm{Cr}^{3} $$ In this reaction, the potassium dichromate is orange and the chromium(III) jons are blue. The bluer a solution of potassium dichromate becomes, the greater the concentration of alcohol in the breath. Write a balanced ionic cquation for this reaction in an acidic cnvironment.

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