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The copper(II) ion in a copper(II) sulfate solution reacts with potassium iodide to produce the triiodide ion, \(\mathrm{I}_{3}^{-}\). This reaction is commonly used to determine how much copper is present in a given sample. $$\operatorname{CuSO}_{4}(a q)+\operatorname{KI}(a q) \rightarrow \operatorname{CuI}(s)+\mathrm{KI}_{3}(a q)+\mathrm{K}_{2} \mathrm{SO}_{4}(a q)$$ If \(2.00 \mathrm{g}\) of \(\mathrm{KI}\) is added to a solution containing 0.525 \(\mathrm{g}\) of \(\mathrm{CuSO}_{4},\) calculate the mass of each product produced.

Short Answer

Expert verified
The mass of each product produced is as follows: \(CuI = 0.626 \, g\), \(KI_3 = 1.29 \, g\), and \(K_2 SO_4 = 0.573 \, g\).

Step by step solution

01

Write the balanced chemical equation

First, write the balanced chemical equation for the reaction: \(CuSO_4(aq) + 2KI(aq) \rightarrow CuI(s) + KI_3(aq) + K_2SO_4(aq)\) This equation tells us the stoichiometric ratios between the reactants and products: 1 mole of \(CuSO_4\) reacts with 2 moles of \(KI\) to produce 1 mole of \(CuI\), 1 mole of \(KI_3\), and 1 mole of \(K_2SO_4\).
02

Determine the limiting reactant

Now, we need to determine which reactant is limiting. Convert the given masses of reactants to moles using their molar masses: For \(CuSO_4\): mass = 0.525 g, molar mass = 159.61 g/mol Moles of \(CuSO_4\) = (0.525 g) / (159.61 g/mol) = 0.00329 mol For \(KI\): mass = 2.00 g, molar mass = 166.0 g/mol Moles of \(KI\) = (2.00 g) / (166.0 g/mol) = 0.0120 mol Now, divide the moles of each reactant by their stoichiometric coefficients, and find the smallest value: \(CuSO_4\): 0.00329 mol / 1 = 0.00329 \(KI\): 0.0120 mol / 2 = 0.00600 Since 0.00329 is smaller than 0.006, \(CuSO_4\) is the limiting reactant.
03

Calculate the mass of each product

Based on the stoichiometric ratios and the limiting reactant (\(CuSO_4\)), we can calculate the mass of each product produced: 1. Mass of \(CuI\) produced: Moles of \(CuI\) = Moles of \(CuSO_4\) (since their stoichiometric ratio is 1:1) = 0.00329 mol Mass of \(CuI\) = moles × molar mass = (0.00329 mol) × (190.45 g/mol) = 0.626 g 2. Mass of \(KI_3\) produced: Moles of \(KI_3\) = Moles of \(CuSO_4\) (since their stoichiometric ratio is 1:1) = 0.00329 mol Mass of \(KI_3\) = moles × molar mass = (0.00329 mol) × (391.82 g/mol) = 1.29 g 3. Mass of \(K_2SO_4\) produced: Moles of \(K_2SO_4\) = Moles of \(CuSO_4\) (since their stoichiometric ratio is 1:1) = 0.00329 mol Mass of \(K_2SO_4\) = moles × molar mass = (0.00329 mol) × (174.26 g/mol) = 0.573 g So, the mass of \(CuI\) produced is 0.626 g, the mass of \(KI_3\) produced is 1.29 g, and the mass of \(K_2SO_4\) produced is 0.573 g.

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Key Concepts

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

Chemical Reaction
A chemical reaction involves the transformation of one set of chemical substances into another. It happens through the breaking and forming of chemical bonds, leading to changes in the composition of matter. The reaction between copper(II) sulfate and potassium iodide to produce triiodide ion is one such chemical transformation.

It's crucial for students to recognize that chemical equations provide a concise way of demonstrating the conversion of reactants to products. The balanced chemical equation for this reaction is fundamental to understanding the stoichiometry, as it dictates the exact proportions in which the substances react and form products.
Limiting Reactant
The concept of the limiting reactant is essential in stoichiometry. It determines the amount of product that can be formed in a chemical reaction. The limiting reactant is the substance that is completely consumed first, which causes the reaction to stop. Once the limiting reactant is depleted, no more product can form, regardless of how much of the other reactants remain.

In the exercise provided, understanding which reactant is limiting involves comparing the mole ratio of the reactants with the coefficients in the balanced equation. By calculating the molar amounts and dividing by the stoichiometric coefficients, students can easily identify the limiting reactant, which in this case is copper(II) sulfate ((CuSO_4)).
Molar Mass
Molar mass, a vital concept in chemistry, is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It directly relates the mass of a substance to the number of moles present, making it indispensable in converting between mass and moles for chemical calculations.

When solving stoichiometry problems, like the copper(II) sulfate-potassium iodide reaction, students must use the molar mass to convert the given mass of each reactant to moles. It's important to extract the correct molar masses from the periodic table for each chemical compound involved in the reaction, as any error in these values can lead to incorrect outcomes in determining the limiting reactant and the mass of products formed.

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Most popular questions from this chapter

Hydrazine, \(\mathrm{N}_{2} \mathrm{H}_{4},\) emits a large quantity of energy when it reacts with oxygen, which has led to hydrazine's use as a fuel for rockets: $$ \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{O}_{2}(g) \rightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ How many moles of each of the gaseous products are produced when \(20.0 \mathrm{g}\) of pure hydrazine is ignited in the presence of \(20.0 \mathrm{g}\) of pure oxygen? How many grams of each product are produced?

What do the coefficients of a balanced chemical equation tell us about the proportions in which substances react on a macroscopic (mole) basis?

The compound sodium thiosulfate pentahydrate, \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3} \cdot 5 \mathrm{H}_{2} \mathrm{O},\) is important commercially to the photography business as "hypo," because it has the ability to dissolve unreacted silver salts from photographic film during development. Sodium thiosulfate pentahydrate can be produced by boiling elemental sulfur in an aqueous solution of sodium sulfite. $$\mathrm{S}_{8}(s)+\mathrm{Na}_{2} \mathrm{SO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3} \cdot 5 \mathrm{H}_{2} \mathrm{O}(s)$$ (unbalanced) What is the theoretical yield of sodium thiosulfate pentahydrate when \(3.25 \mathrm{g}\) of sulfur is boiled with 13.1 g of sodium sulfite? Sodium thiosulfate pentahydrate is very soluble in water. What is the percent yield of the synthesis if a student doing this experiment is able to isolate (collect) only \(5.26 \mathrm{g}\) of the product?

If sodium peroxide is added to water, elemental oxygen gas is generated: $$\mathrm{Na}_{2} \mathrm{O}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{NaOH}(a q)+\mathrm{O}_{2}(g)$$ Suppose \(3.25 \mathrm{g}\) of sodium peroxide is added to a large excess of water. What mass of oxygen gas will be produced?

Explain how one determines which reactant in a process is the limiting reactant. Does this depend only on the masses of the reactant present? Is the mole ratio in which the reactants combine involved?

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