Salt Analysis
Reagent S is
The precipitate P contains
The coloured solution S contains
The compound N is
The final solution contains :
Passing H2S gas into a mixture of Mn2+, Ni2+, Cu2+ and Hg2+ ions in an acidified aqueous solution precipitates
The compound X is
The compound Y is
The compound Z is
A solution of a metal ion when treated with KI gives a red precipitate which dissolves in excess KI to give a colourless solution. Moreover, the solution of metal ion treatment with a solution of cobalt (II) thiocyanate gives rise to a deep blue crystalline precipitate. The metal ion is:
A solution when diluted with $\mathrm{H}_2 \mathrm{O}$ and boiled, it gives a white precipitate. On addition of excess $\mathrm{NH}_4 \mathrm{Cl} / \mathrm{NH}_4 \mathrm{OH}$, the volume of precipitate decreases leaving behind a white gelatinous precipitate. Identify the precipitate which dissolves in $\mathrm{NH}_4 \mathrm{OH} / \mathrm{NH}_4 \mathrm{Cl}$.
$\mathrm{Zn}(\mathrm{OH})_2$
$\mathrm{Al}(\mathrm{OH})_3$
$\mathrm{Mg}(\mathrm{OH})_2$
$\mathrm{Ca}(\mathrm{OH})_2$
$\mathrm{CuSO}_4$ decolourises on addition of KCN , the product is
$\left[\mathrm{Cu}(\mathrm{CN})_4\right]^{2-}$
$\mathrm{Cu}^{2+}$ get reduced to form $\left[\mathrm{Cu}(\mathrm{CN})_4\right]^{3-}$
$\mathrm{Cu}(\mathrm{CN})_2$
CuCN
Explanation:
Univalent metal means alkali metal. A is binary compound so A can be KO or KO2.
But here A react completely with sulphur.
$\therefore$ A must be KO2.
To check A and B are right or not we have to check with their molar ratio.
Here molar ratio of KO2 and S = 2 : 1
Moles of 1.422 g KO2 = ${{1.422} \over {71}} = 0.02$
Moles of 0.321 g S = ${{0.321} \over {32}} = 0.01$
$\therefore$ Molar ratio = 0.02 : 0.01 = 2 : 1
$\therefore$ A and B are right choice.
When K+ is the monovalent cation and Al+3 is the trivalent cation then double salt C is $ = {K_2}S{O_4}.A{l_2}{(S{O_4})_3}.24{H_2}O$
When 16.8 g of white solid X were heated, 4.4 g of acid gas A, that turned lime water milky was driven off together with 1.8 g of a gas B which condensed to a colourless liquid.
The solid that remained, Y, dissolved in water to give an alkaline solution, which with excess barium chloride solution gave a white precipitate Z. The precipitate efferversced with acid giving off carbon dioxide. Identify A, B and Y and write down the equation for the thermal decomposition of X.
Explanation:
(i) Gas A turned lime water milky, which indicates the presence of carbon dioxide. Therefore, gas A is carbon dioxide (CO2).
(ii) Gas B, which condenses to a colourless liquid: This is most likely water (H2O).
(iii) $Y$ when dissolved in water yields an alkaline solution and the solution on treatment with $\mathrm{BaCl}_2$ solution forms a white ppt, of $Z$. The compound $Z$ on treatment with acid gives effervescence of $\mathrm{CO}_2$ so $Z$ and hence $Y$ must be a carbonate, $\mathrm{CO}_3^{2-}$. We can thus write $Y$ as $M \mathrm{CO}_3$ or $M_2 \mathrm{CO}_3$.
(iv) $X$ on being heated yields a carbonate $Y$, hence $\mathrm{CO}_{2(g)}$ i.e. $A$ and another gas $B$, hence it must be a bicarbonate, $\mathrm{HCO}_3^{-}$.
$4.4 \mathrm{~g}$ of $\mathrm{CO}_2$ is given by $M\left(\mathrm{HCO}_3\right)=16.8 \mathrm{~g}$
$44 \mathrm{~g}$ of $\mathrm{CO}_2$ is given by $M\left(\mathrm{HCO}_3\right)$
$ =\frac{16.8}{4.4} \times 44=168 \mathrm{~g} $
Because 2 molecules of $M\left(\mathrm{HCO}_3\right)$ are involved in the reaction so molecular weight of
$M\left(\mathrm{HCO}_3\right)=168 / 2=84$
Let the atomic weight be $M$.
Molecular weight of $M\left(\mathrm{HCO}_3\right)=M+1+12+48$$=M+61$
$ \therefore M+61=84 \text { or } M=84-61=23 $
Thus the metal must be $\mathrm{Na}$ and so the given salt $X$ is $\mathrm{Na}\left(\mathrm{HCO}_3\right)$.
Explain the following in not more than two sentenses.
A solution of FeCl3 in water gives a brown precipitate on standing.
Explanation:
$ \mathrm{FeCl}_3+3 \mathrm{H}_2 \mathrm{O} \longrightarrow 3 \mathrm{HCl}+\mathrm{Fe}(\mathrm{OH})_3 \downarrow $(Brown)
Compound A is a light green crystalline solid. It gives the following tests :
(i) It dissolves in dilute sulphuric acid. No gas is produced.
(ii) A drop of KMnO4 is added to the above solution. The pink colour disappears.
(iii) Compound A is heated strongly. Gases B and C, with pungent smell, come out. A brown residue D is left behind.
(iv) The gas mixture (B and C) is passed into a dichromate solution. The solution turns green.
(v) The green solution from step (iv) gives a white precipitate E with a solution of barium nitrate.
(vi) Residue D from step (iii) is heated on charcoal in a reducing flame. It gives a magnetic substance E.
Name the compounds A, B, C, D and E.
Explanation:
Let's analyze the given information step-by-step to identify the compounds.
(i) Dissolution in dilute sulfuric acid without gas evolution: This suggests that compound A is likely a metal compound that forms a soluble sulfate. The lack of gas rules out carbonates and many other reactive compounds.
(ii) Decolorization of KMnO₄: Potassium permanganate ($KMnO_4$) is a strong oxidizing agent. Its decolorization indicates that compound A, or a species derived from it in the acidic solution, is readily oxidized. This points towards a compound containing a metal in a lower oxidation state, capable of being oxidized to a higher oxidation state.
(iii) Strong heating producing pungent gases (B and C) and brown residue (D): The pungent gases suggest the presence of sulfur and/or nitrogen. The brown residue is indicative of a metal oxide. The fact that gases are produced implies that compound A might be a metal salt containing an anion which decomposes upon heating.
(iv) Gases B and C turn a dichromate solution green: Dichromate ($Cr_2O_7^{2-}$) solutions are orange. Their reduction to a green solution indicates that gases B and C are reducing agents. The green color is characteristic of chromium(III) ions ($Cr^{3+}$). This strongly suggests that one of the gases (B or C) is $SO_2$, which is capable of reducing dichromate.
(v) The green solution gives a white precipitate (E) with barium nitrate: The formation of a white precipitate with barium nitrate is a classic test for sulfate ions ($SO_4^{2-}$). The reaction is:
$Ba^{2+}(aq) \ + \ SO_4^{2-}(aq) \ \rightarrow \ BaSO_4(s)$
Therefore, gas B or C must be sulfur dioxide ($SO_2$), which is oxidized to sulfate ions ($SO_4^{2-}$) upon reaction with the dichromate solution.
(vi) Residue D gives a magnetic substance (E) upon reduction: This strongly suggests that residue D is a metal oxide of a transition metal, and the magnetic substance E is the corresponding metal. The reduction reaction can be generally represented as:
$Metal Oxide \ + \ Reducing \ Agent \ \rightarrow \ Metal$
Considering all the observations, a plausible scenario is that compound A is ferrous sulfate heptahydrate ($FeSO_4 \cdot 7H_2O$). This is a light green crystalline solid. Let's check if it fits the observations:
(i) & (ii): Ferrous sulfate dissolves in dilute sulfuric acid and gets oxidized by $KMnO_4$.
(iii): Upon strong heating, ferrous sulfate decomposes. The gases produced include sulfur dioxide ($SO_2$ - gas B) and sulfur trioxide ($SO_3$ - gas C). Both have pungent smells. The brown residue (D) is ferric oxide ($Fe_2O_3$).
(iv): $SO_2$ reduces dichromate to $Cr^{3+}$.
(v): $SO_2$ is oxidized to $SO_4^{2-}$, which forms a white precipitate of barium sulfate ($BaSO_4$) with barium nitrate.
(vi): Reduction of $Fe_2O_3$ on charcoal produces elemental iron (Fe - E), which is magnetic.
Therefore, the compounds are:
A: Ferrous sulfate heptahydrate ($FeSO_4 \cdot 7H_2O$)
B: Sulfur dioxide ($SO_2$)
C: Sulfur trioxide ($SO_3$)
D: Ferric oxide ($Fe_2O_3$)
E: Iron (Fe) (or Barium Sulfate ($BaSO_4$), depending on which step (v) or (vi) is considered)
A white amorphous powder (A) on heating yields a colourless, non-combustible 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 K4Fe(CN)6 solution.
(A) dissolves in dilute HCl with the evolution of gas, which is identical in all respects with (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 NH4OH and passing H2S. Another portion of the solution gives initially a white precipitate (E) on the addition of sodium hydroxide solution, which dissolves on further addition of the base. Identify the compounds (A), (B), (C), (D) and (E).
Explanation:
$ \begin{aligned} & A \stackrel{\Delta}{\longrightarrow} { } \underset{C}{\mathrm{ZnO}}+B \text { (gas) } \\\\ & \underset{C} {\mathrm{ZnO}}+2 \mathrm{HCl} \longrightarrow \mathrm{ZnCl}_2+\mathrm{H}_2 \mathrm{O} \\\\ & 2 \mathrm{ZnCl}_2+\mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right] \longrightarrow 4 \mathrm{KCl}+\mathrm{Zn}_2\left[\mathrm{Fe}(\mathrm{CN})_6\right] \downarrow (White ppt) \end{aligned} $
(ii) The gas $B$ turns lime water milky and milkiness disappear with continuous passage of gas. Hence, the gas is $\mathrm{CO}_2$ and compound $A$ in $\mathrm{ZnCO}_3$.
$ \begin{aligned} \underset{C}{\mathrm{CO}_2}+\mathrm{Ca}(\mathrm{OH})_2 & \longrightarrow \mathrm{H}_2 \mathrm{O}+\mathrm{CaCO}_3 \downarrow \\\\ \mathrm{CaCO}_3+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O} & \longrightarrow \mathrm{Ca}\left(\mathrm{HCO}_3\right)_2 \\\\ \underset{A} {\mathrm{ZnCO}_3} & \stackrel{\Delta}{\longrightarrow} \underset{C}{\mathrm{ZnO}}+\underset{B}{\mathrm{CO}_2} \end{aligned} $
(iii) The solution of $A$ gives white ppt of $\mathrm{ZnS} D$ with $\mathrm{NH}_4 \mathrm{OH}$ and excess of $\mathrm{H}_2 \mathrm{~S}$.
$ \begin{aligned} & \mathrm{ZnCO}_3+\mathrm{HCl} \longrightarrow \underset{C}{\mathrm{CO}_2 \uparrow}+\mathrm{ZnCl}_2 \\\\ & \mathrm{ZnCl}_2+\mathrm{H}_2 \mathrm{~S} \stackrel{\mathrm{NH}_4 \mathrm{OH}}{\longrightarrow} 2 \mathrm{HCl}+{\underset{D}{\mathrm{ZnS}} \downarrow \text { (white) }} \end{aligned} $
(iv) The solution of $A$ also gives initially a white ppt $E$ with $\mathrm{NaOH}$, which dissolve in excess of reagent.
$ \begin{aligned} & \mathrm{ZnCl}_2+2 \mathrm{NaOH} \longrightarrow \underset{E \text { (white) }} {\mathrm{Zn}(\mathrm{OH})_2} \downarrow+2 \mathrm{NaCl} \\\\ & \mathrm{Zn}(\mathrm{OH})_2+2 \mathrm{NaOH} \longrightarrow \underset{ \text { (Soluable) }}{\mathrm{Na}_2\left[\mathrm{Zn}(\mathrm{OH})_4\right]} \end{aligned} $