Question

A substance \(X\) when heated with \(Y\) produces residue and odourless, gas which turns lime water milky. The silver nitrate is added to residue yellow precipitate is formed. \(X\) and \(Y\) could be \(\begin{array}{ll}\mathbf{X} & \mathbf{Y}\end{array}\) \(\begin{array}{ll}\mathbf{X} & \mathbf{Y}\end{array}\) (1) \(\mathrm{MnO}_{2} \mathrm{~S}\) (2) \(\mathrm{NaBrO}_{3} \mathrm{C}\) (3) \(\mathrm{Kl} \quad \mathrm{S}\) (4) \(\mathrm{Na}_{2} \mathrm{SO}_{4} \mathrm{C}\)

Short answer

(3) KI and S

Step-by-step solution

Identify the Gas

The problem states that a gas is produced which turns lime water milky. This indicates the gas is likely carbon dioxide \((CO_2)\), as it forms calcium carbonate \((CaCO_3)\) when bubbled through lime water.

Analyze Reactions with Silver Nitrate

The residue forms a yellow precipitate with silver nitrate. Silver nitrate \(AgNO_3\) reacts with certain substances to form distinct precipitates: for instance, Iodides (like \(KI\)) form yellow precipitates of silver iodide \((AgI)\).

Consider Possible Combinations

Given the observations, the likely combinations for \(X\) and \(Y\) are: (1) \(MnO_2 \text{and} \; S\) (2) \(NaBrO_3 \text{and} \; C\) (3) \(KI \text{and} \; S\) (4) \(Na_2SO_4 \text{and} \; C\).

Eliminate Impossible Combinations

We can eliminate options where the reaction does not produce a gas that turns lime water milky (i.e., does not produce \(CO_2\)). For example: - \(MnO_2\) with \(S\) does not produce \(CO_2\) - \(Na_2SO_4\) with \(C\) is also unlikely to produce \(CO_2\).

Confirm the Correct Combination

The combination of \(KI\) and \(S\) is plausible, as \(KI\) could react to produce a residue that forms a yellow precipitate with \ AgNO_3 \, and \(S\) (sulfur) can produce gases when heated.

Key concepts

reaction identification

Identifying a chemical reaction is key to solving chemistry problems. To do this, observe the products formed from the reactants. In the given problem, when the substance X reacts with Y, it produces a residue and a gas that turns lime water milky. This description helps us understand which chemicals are involved.

The gas that turns lime water milky is a common clue, indicating the presence of carbon dioxide \(CO_2\). This gas forms when substances containing carbon are burnt or decomposed. For example, if a substance containing carbon is heated, it often releases \(CO_2\), which reacts with lime water \(Ca(OH)_2\) to form calcium carbonate \(CaCO_3\), making the water turn milky.

Understanding reaction identification helps you focus and narrow down possible substances involved. Keep an eye on such notable changes to decipher the chemical processes easily.

silver nitrate test

The silver nitrate test is a standard method to identify halides in a compound. When silver nitrate \(AgNO_3\) is added to a solution containing halide ions, it forms a precipitate. The color of the precipitate can indicate which halide is present.

- Chlorides (\(Cl^-\)) form white precipitates of silver chloride (\(AgCl\)).
- Bromides (\(Br^-\)) form cream-colored precipitates of silver bromide (\(AgBr\)).
- Iodides (\(I^-\)) form yellow precipitates of silver iodide (\(AgI\)).

In the given problem, adding silver nitrate to the residue forms a yellow precipitate. This suggests that the residue contains iodide ions. Thus, the substance X is likely to be potassium iodide (KI), which, upon heating, can leave a residue that reacts with silver nitrate to form yellow \(AgI\).

Using well-known tests like the silver nitrate test can provide significant clues about the compounds you are analyzing.

carbon dioxide detection

Detecting carbon dioxide is often done through its reaction with lime water. When carbon dioxide is bubbled through lime water \(Ca(OH)_2\), it forms calcium carbonate \(CaCO_3\), which is a milky white precipitate.

This reaction is helpful for several reasons:

- It's a quick and efficient method to confirm the presence of \(CO_2\) in a sample.
- It helps identify the substance being heated or reacted, especially if \(CO_2\) is a known product.

In the problem, carbon dioxide detection helped identify that one product from heating the substances was a gas turning lime water milky, hinting at \(CO_2\).

Understanding this simple, yet powerful chemical property aids in identifying gases in numerous experimental chemistry scenarios.

chemical combinations

Understanding chemical combinations is crucial in predicting the outcome of a reaction. Looking at the reactants and products can help determine which combinations are plausible.

In the given exercise, several combinations were provided: \text{(MnO}_2\text{ & S), (NaBrO}_3\text{ & C), (KI & S), (Na}_2\text{SO}_4\text{ & C)}\. Each pair would undergo different reactions to form distinct products.

Key points to consider:

• The elemental composition of reactants, especially in heating scenarios, to predict resulting products.
• Knowledge of chemical properties, such as whether a substance will produce \(CO_2\) or react with silver nitrate to form a precipitate.

Through elimination and understanding properties, we determined \(KI \text{ & } S\) is the correct combination. The iodide reacts to form a yellow \(AgI\) precipitate, and heating sulfur-containing compounds often produces \(CO_2\).

Mastering chemical combinations and their likely outcomes aids immensely in problem-solving in chemistry.