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State with balanced equations, what happens when (i) Silver is treated with hot concentrated sulphuric acid. (ii) Ammonium dichromate is heated. (iii) Hydrogen sulphide is passed through a solution of potassium permanganate acidified with dilute sulphuric acid.

Short Answer

Expert verified
1) Silver reacts with concentrated H₂SO₄ to form Ag₂SO₄ and SO₂. 2) Ammonium dichromate decomposes into Cr₂O₃, N₂, and H₂O. 3) H₂S with KMnO₄ in acidic solution forms MnSO₄, sulfur, and water.

Step by step solution

01

Reaction of Silver with Hot Concentrated Sulfuric Acid

When silver (Ag) reacts with hot concentrated sulfuric acid (Hâ‚‚SOâ‚„), silver sulfate (Agâ‚‚SOâ‚„) is formed along with sulfur dioxide (SOâ‚‚) gas, and water (Hâ‚‚O). The balanced chemical equation for this reaction is:\[2Ag + 2H_2SO_4 (conc) ightarrow Ag_2SO_4 + 2H_2O + SO_2 \]
02

Heating Ammonium Dichromate

When ammonium dichromate \((NH_4)_2Cr_2O_7\) is heated, it decomposes to produce chromium(III) oxide (Cr₂O₃), nitrogen gas (N₂), and water vapor (H₂O). The balanced chemical equation is:\[(NH_4)_2Cr_2O_7 ightarrow Cr_2O_3 + N_2 + 4H_2O\]
03

Reaction of Hydrogen Sulfide with Potassium Permanganate in Acidic Solution

When hydrogen sulfide (Hâ‚‚S) is passed through a solution of potassium permanganate (KMnOâ‚„) acidified with dilute sulfuric acid (Hâ‚‚SOâ‚„), manganese sulfate (MnSOâ‚„), sulfur (S), and water (Hâ‚‚O) are produced. The balanced chemical equation is:\[2KMnO_4 + 3H_2S + 2H_2SO_4 ightarrow 2MnSO_4 + K_2SO_4 + 3S + 4H_2O\]

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Key Concepts

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

Redox Reactions
In the realm of chemical reactions, redox reactions—or oxidation-reduction reactions—hold a significant place. These reactions involve the transfer of electrons between two species.

The term "redox" comes from the two fundamental processes occurring simultaneously:
  • Oxidation: The loss of electrons by a molecule, atom, or ion.
  • Reduction: The gain of electrons by a molecule, atom, or ion.
In the example of hydrogen sulfide (\( \text{H}_2\text{S} \)) reacting with potassium permanganate (\( \text{KMnO}_4 \)), we observe a redox reaction. Here, hydrogen sulfide is oxidized to sulfur, as it loses electrons (oxidation), and the manganese in potassium permanganate is reduced to manganese sulfate as it gains electrons (reduction).

It is crucial to identify both oxidation and reduction processes in any redox reaction as they always occur together. Redox reactions are balanced not just for atoms but also for charge. Make sure electrons gained and lost are the same in a perfectly balanced redox equation.
Stoichiometry
Stoichiometry is the mathematical backbone of chemistry, providing us with the tools to quantify relationships in chemical reactions. It helps us determine the exact proportions of reactants and products involved to ensure a balanced and correct reaction equation.

When silver reacts with sulfuric acid, stoichiometry is used to balance the equation:\[2\text{Ag} + 2\text{H}_2\text{SO}_4 \rightarrow \text{Ag}_2\text{SO}_4 + 2\text{H}_2\text{O} + \text{SO}_2\]To facilitate this, one must ensure that the number of atoms for each element is equal on both sides of the reaction equation. Stoichiometry involves working with coefficients that represent moles, ensuring the law of conservation of mass is adhered to, meaning no atoms are lost or gained in the process.

Practical steps in stoichiometry include:
  • Identifying and writing down the chemical reactions, making sure they're correctly balanced.
  • Calculating the moles of reactants and products using their molar masses.
  • Using mole ratios derived from the balanced equation to find out how much of each substance is required or produced.
Chemical Equations
Chemical equations are the symbolic representation of chemical reactions. They provide a concise way to express what happens when substances transform into different substances.

Each chemical equation consists of reactants—the starting materials—and products—the resulting substances. Writing a chemical equation involves not only knowing the substances involved but also how they interact, break apart, or come together to form new compounds.For example, when ammonium dichromate (\((NH_4)_2Cr_2O_7\)) decomposes upon heating, it forms chromium(III) oxide (\(Cr_2O_3\)), nitrogen gas (\(N_2\)), and water (\(H_2O\)). The balanced chemical equation is given as:\[(NH_4)_2Cr_2O_7 \rightarrow Cr_2O_3 + N_2 + 4H_2O\]

A well-formed chemical equation must be balanced:
  • This means the number of each type of atom must be the same on both sides of the equation.
  • It must respect the conservation of mass principle.
Balanced equations are essential for stoichiometry, helping to quantify reactants' measures and products in a chemical process.

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Most popular questions from this chapter

Thermal decomposition of a Mn compound (X) at \(513 \mathrm{~K}\) results in compound \(\mathrm{Y}, \mathrm{MnO}_{2}\) and a gaseous product. \(\mathrm{MnO}_{2}\) reacts with \(\mathrm{NaCl}\) and concentrated \(\mathrm{H}_{2} \mathrm{SO}_{4}\) to give a pungent gas \(\mathrm{Z} . \mathrm{X}, \mathrm{Y}\), and \(\mathrm{Z}\), respectively, are: (a) \(\mathrm{KMnO}_{4}, \mathrm{~K}_{2} \mathrm{MnO}_{4}\) and \(\mathrm{Cl}_{2}\) (b) \(\mathrm{K}_{2} \mathrm{MnO}_{4}, \mathrm{KMnO}_{4}\) and \(\mathrm{SO}_{2}\) (c) \(\mathrm{K}_{3} \mathrm{MnO}_{4}, \mathrm{~K}_{2} \mathrm{MnO}_{4}\) and \(\mathrm{Cl}_{2}\) (d) \(\mathrm{K}_{2} \mathrm{MnO}_{4}, \mathrm{KMnO}_{4}\) and \(\mathrm{Cl}_{2}\)

A white amorphous powder (A) on heating yields a colourless, noncombustible gas (B) and a solid (C). The latter compound assumes a yellow colour on heating and changes to white on cooling. ' \(C\) ' dissolves in dilute acid and the resulting solution gives a white precipitate on adding \(\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]\) solution. A' dissolves in dilute HCl with the evolution of gas, which is identical in all respects with ' \(\mathrm{B}\) '. The gas 'B' turns lime water milky, but the milkiness disappears with the continuous passage of gas. The solution of ' \(A\) ', as obtained above, gives a white precipitate (D) on the addition of excess of \(\mathrm{NH}_{4} \mathrm{OH}\) and passing \(\mathrm{H}_{2} \mathrm{~S}\). Another portion of the solution gives initially a white precipitate (E) on the addition of sodium hydroxide solution, which dissolves on futher addition of the base. Identify the compounds \(\mathrm{A}, \mathrm{B}, \mathrm{D}\), and \(\mathrm{E}\).

The product of oxidation of \(\mathrm{I}^{-}\)with \(\mathrm{MnO}_{4}^{-}\)in alkaline medium is (a) \(\mathrm{IO}_{3}^{-}\) (b) \(\mathrm{I}_{2}\) (c) IO- (d) \(\mathrm{IO}_{4}^{-}\)

An aqueous solution of \(\mathrm{FeSO}_{4}, \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) and chrome alum is heated with excess of \(\mathrm{Na}_{2} \mathrm{O}_{2}\) and filtered. The materials obtained are : (a) a colourless filtrate and a green residue (b) a yellow filtrate and a green residue (c) a yellow filtrate and a brown residue (d) a green filtrate and a brown residue

Which of the following is the weakest base (a) \(\mathrm{NaOH}\) (b) \(\mathrm{Ca}(\mathrm{OH})_{2}\) (c) \(\mathrm{KOH}\) (d) \(\mathrm{Zn}(\mathrm{OH})_{2}\)

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