/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 135 The unknown acid \(\mathrm{H}_{2... [FREE SOLUTION] | 91影视

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The unknown acid \(\mathrm{H}_{2} \mathrm{X}\) can be neutralized completely by \(\mathrm{OH}^{-}\) according to the following (unbalanced) equation: $$ \mathrm{H}_{2} \mathrm{X}(a q)+\mathrm{OH}^{-} \longrightarrow \mathrm{X}^{2-}+\mathrm{H}_{2} \mathrm{O} $$ The ion formed as a product, \(\mathrm{X}^{2-}\), was shown to have 36 total electrons. What is element \(\mathrm{X}\) ? Propose a name for \(\mathrm{H}_{2} \mathrm{X}\). To completely neutralize a sample of \(\mathrm{H}_{2} \mathrm{X}, 35.6 \mathrm{~mL}\) of \(0.175 \mathrm{M} \mathrm{OH}^{-}\) solution was required. What was the mass of the \(\mathrm{H}_{2} \mathrm{X}\) sample used?

Short Answer

Expert verified
The element X is Selenium (Se), and the compound H鈧俋 is called hydrogen selenide or selenous acid. The mass of the H鈧俋 sample used for neutralization is 0.252 g.

Step by step solution

01

Identify Element X

Since the element X forms an ion X虏鈦 with 36 electrons, we can find the number of protons (atomic number) in the neutral atom of X. The ion has 2 extra negative charges (electrons), so the neutral atom has 36 - 2 = 34 electrons, which equals the number of protons. Element X is Selenium (Se), which has an atomic number of 34.
02

Provide a Name for H鈧俋

With the element X identified as Selenium, we can now name the unknown acid H鈧俋. Since it is an acid with the Se虏鈦 ion, it is called hydrogen selenide or selenous acid.
03

Calculate Moles of OH鈦

To find the moles of OH鈦 needed for neutralization, we can use the volume and concentration of the OH鈦 solution: moles of OH鈦 = volume 脳 concentration moles of OH鈦 = 35.6 mL 脳 0.175 mol/L = 6.23 脳 10鈦宦 mol
04

Calculate Moles of H鈧俋

Using the balanced neutralization equation: H鈧俋 + 2 OH鈦 鈫 X虏鈦 + 2 H鈧侽 The mole ratio between H鈧俋 and OH鈦 is 1:2, so we can find the moles of H鈧俋 as follows: moles of H鈧俋 = 1/2 脳 moles of OH鈦 = 1/2 脳 6.23 脳 10鈦宦 mol = 3.115 脳 10鈦宦 mol
05

Determine the Mass of H鈧俋 Sample

To find the mass of the H鈧俋 sample, we first need to calculate the molar mass of H鈧俋 (H鈧係e): M(H鈧係e) = 2 脳 M(H) + M(Se) = 2 脳 1.008 g/mol + 78.96 g/mol = 80.976 g/mol Now, we can calculate the mass of the H鈧俋 sample as follows: mass of H鈧俋 = moles of H鈧俋 脳 molar mass of H鈧係e mass of H鈧俋 = 3.115 脳 10鈦宦 mol 脳 80.976 g/mol = 0.252 g The mass of the H鈧俋 sample used is 0.252 g.

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Key Concepts

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

Chemical Nomenclature
Chemical nomenclature provides a language for chemists to communicate about substances clearly and precisely. In the given problem, the substance \(\mathrm{H}_2 \mathrm{X}\) is an unknown acid. Once the element X is recognized as Selenium (Se), naming the compound becomes straightforward. \(\mathrm{H}_2 \mathrm{X}\), with selenium as the central element, forms the compound hydrogen selenide when hydrogen ions are added. If the compound stays in a lower oxidation state, it would primarily be called hydrogen selenide, but as an acid, it's typically referred to as selenous acid.
  • The rules of chemical nomenclature maintain that binary acids, comprising hydrogen and a nonmetal, take the prefix 'hydro-' with the anion root name followed by '-ic' for most cases. However, 'selenous acid' follows different historical standards for ternary acids.
  • Nomenclature not only aids identification but falls in line with agreed structural (oxidation states) rules.
Understanding these naming conventions helps students quickly deduce the composition and properties of compounds in chemistry.
Stoichiometry
Stoichiometry pertains to the calculations based on balanced chemical reactions. In neutralization reactions, this concept is particularly important as it helps determine the actual amounts needed for complete reaction. The equation considered is: \[\mathrm{H}_2\mathrm{X} + 2\mathrm{OH}^- \rightarrow \mathrm{X}^{2-} + 2\mathrm{H}_2\mathrm{O}\]
  • From the balanced reaction, 1 mole of \(\mathrm{H}_2 \mathrm{X}\) reacts with 2 moles of \(\mathrm{OH}^-\) ions.
  • Steps to balance a chemical equation typically involve equating the number of atoms of each element on both sides of the reaction.
Stoichiometry allows the conceptual accounting of reactants and products, ensuring that in this chemical scenario, no atoms are lost but metamorphosed into new bonds and molecules.
Molar Mass Calculation
Calculating molar mass involves summing the atomic masses of all atoms that make up a given molecule. With \(\mathrm{H}_2\mathrm{Se}\) identified, we determine its molar mass:
\[M(\mathrm{H}_2\mathrm{Se}) = 2 \times M(\mathrm{H}) + M(\mathrm{Se}) = 2 \times 1.008 \, \text{g/mol} + 78.96 \, \text{g/mol} = 80.976 \, \text{g/mol}\]
  • The atomic mass of Hydrogen is approximately 1.008 g/mol, while Selenium's is 78.96 g/mol.
  • Molar mass is essentially the weight of one mole of a substance, crucial for converting between the mass of a sample and the quantity in moles.
Understanding molar mass calculations is vital to bridging comprehensive mass-based lab measurements and stoichiometric calculations.
Acid-Base Titration
Acid-base titration is a lab technique we use to determine the concentration of an unknown acid or base by reacting it with a titrant of known concentration. In this problem, the titration of \(\mathrm{H}_2\mathrm{X}\) with \(\mathrm{OH}^-\) finds how much base is needed to completely neutralize the acid.
  • By using \(35.6 \, \text{mL} \, \text{of} \, 0.175 \, \text{M} \, \mathrm{OH}^-\) solution, the moles of hydroxide ions are found by multiplying its concentration by volume.
  • This concept translates into generating a stoichiometry pathway to balance the reaction and calculate the \(\mathrm{H}_2\mathrm{X}\) moles via the 1:2 relationship.
Titration results in easily handling unknown concentrations of substances and forms practical hands-on uses for overarching chemical reactions.

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Most popular questions from this chapter

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