To solve this problem, we need to determine the number of moles of O3 and O2 involved when 1 mole of PbS is oxidized. The chemical reaction for oxidizing PbS using O3 is:
PbS + 4O3 → PbSO4 + 4O2
Step-by-step explanation:
From the balanced reaction, 1 mole of PbS requires 4 moles of O3 to be oxidized to PbSO4.
During this process, it produces 4 moles of O2.
Given these stoichiometric relationships, the number of moles of O3 (“X” moles) used is 4, and the number of moles of O2 (“Y” moles) produced is also 4.
The problem asks for the sum X + Y, which is:
X + Y = 4 + 4 = 8
Verification:
The computed value of X + Y = 8 falls within the expected range of [8,8].
The balanced chemical equation for the oxidation of PbS by ozone is:
\(\text{PbS} + 4\text{O}_3 \rightarrow \text{PbSO}_4 + 4\text{O}_2\)
From the equation:
- 1 mole of PbS reacts with 4 moles of \( \text{O}_3 \), so \( X = 4 \).
- 4 moles of \( \text{O}_2 \) are produced, so \( Y = 4 \).
Therefore:
\(X + Y = 4 + 4 = 8\)
The Correct answer is: 8
Given below are two statements:
Statement (I): The first ionization energy of Pb is greater than that of Sn.
Statement (II): The first ionization energy of Ge is greater than that of Si.
In light of the above statements, choose the correct answer from the options given below:
The product (A) formed in the following reaction sequence is:

\[ \begin{array}{|c|c|} \hline \textbf{LIST-I (Redox Reaction)} & \textbf{LIST-II (Type of Redox Reaction)} \\ \hline A. \, CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l) & I. \, \text{Disproportionation reaction} \\ B. \, 2NaH(s) \rightarrow 2Na(s) + H_2(g) & II. \, \text{Combination reaction} \\ C. \, V_2O_5(s) + 5Ca(s) \rightarrow 2V(s) + 5CaO(s) & III. \, \text{Decomposition reaction} \\ D. \, 2H_2O(aq) \rightarrow 2H_2(g) + O_2(g) & IV. \, \text{Displacement reaction} \\ \hline \end{array} \]
Let one focus of the hyperbola \( H : \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1 \) be at \( (\sqrt{10}, 0) \) and the corresponding directrix be \( x = \dfrac{9}{\sqrt{10}} \). If \( e \) and \( l \) respectively are the eccentricity and the length of the latus rectum of \( H \), then \( 9 \left(e^2 + l \right) \) is equal to:
