Match the reactions (in the given stoichiometry of the reactants) in List-I with one of their products given in List-II and choose the correct option.
List-I
List-II
(P) P2O3 + 3H2O → (1) P(O)(OCH3)Cl2 (Q) P4 + 3NaOH + 3H2O → (2) H3PO3 (R) PCl5 + CH3COOH → (3) PH3 (S) H3PO2 + 2H2O + 4AgNO3 → (4) POCl3 (5) H3PO4
Match the reactions (in the given stoichiometry of the reactants) in List-I with one of their products given in List-II and choose the correct option.
List-I | List-II | ||
| (P) | P2O3 + 3H2O → | (1) | P(O)(OCH3)Cl2 |
| (Q) | P4 + 3NaOH + 3H2O → | (2) | H3PO3 |
| (R) | PCl5 + CH3COOH → | (3) | PH3 |
| (S) | H3PO2 + 2H2O + 4AgNO3 → | (4) | POCl3 |
| (5) | H3PO4 |
P →2; Q→ 3; R →1; S →5
P →3; Q→ 5; R →4; S →2
P →5; Q→ 2; R →1; S →3
P →2; Q→ 3; R →4; S →5
The Correct Option is D
Solution and Explanation
To solve this problem, we need to match each reaction in List-I with one of their products from List-II based on chemical knowledge.
The reactions are identified by specific chemical behaviors and properties:
- (P) P2O3 + 3H2O →
P2O3 is phosphorus(III) oxide, which reacts with water to form phosphorous acid. Thus, the product is H3PO3 (Phosphorous acid) corresponding to List-II's option (2). - (Q) P4 + 3NaOH + 3H2O →
This reaction is characteristic of white phosphorus reacting with a base (NaOH) and water to produce phosphine gas (PH3) and sodium hypophosphite. Therefore, the product matches PH3, which corresponds to List-II's option (3). - (R) PCl5 + CH3COOH →
Phosphorus pentachloride reacts with acetic acid to form acetyl chloride and Phosphorus oxychloride (POCl3), corresponding to List-II's option (4). - (S) H3PO2 + 2H2O + 4AgNO3 →
Hypophosphorous acid (H3PO2) in presence of water and silver nitrate yields phosphoric acid along with silver and nitric acid. The product is H3PO4 (Phosphoric acid), corresponding to List-II's option (5).
Hence, the correct matching is:
- P → 2
- Q → 3
- R → 4
- S → 5
Top Questions on Stoichiometry and Stoichiometric Calculations
- 1 mole of an ideal gas is allowed to expand isothermally and reversibly from 1L to 5L at 300 K. The change in enthalpy (in kJ) is
(R = 8.3 J K\(^{-1}\) mol\(^{-1}\))- AP EAPCET - 2025
- Chemistry
- Stoichiometry and Stoichiometric Calculations
- The number of extensive and intensive properties in the list given below is respectively: % List - Density, enthalpy, mass, temperature, volume, pressure
- AP EAPCET - 2025
- Chemistry
- Stoichiometry and Stoichiometric Calculations
- The number of moles of CO$_2$ produced when 2 moles of C$_2$H$_6$ are completely burnt is:
- BITSAT - 2025
- Chemistry
- Stoichiometry and Stoichiometric Calculations
- For which reaction \( \Delta H \neq \Delta U \)?
- AP EAPCET - 2025
- Chemistry
- Stoichiometry and Stoichiometric Calculations
- Consider the following reaction A(g) + 3B(g) \( \longrightarrow \) 2C(g); \( \Delta H^\ominus = -24 \) kJ. At \(25 \, ^\circ\text{C}\) if \( \Delta G^\ominus \) of the reaction is -9 kJ, the standard entropy change (in JK\(^{-1}\)) of the same reaction at same temperature is
- AP EAPCET - 2025
- Chemistry
- Stoichiometry and Stoichiometric Calculations
Questions Asked in JEE Advanced exam
- At 25$^\circ$C, the concentration of H$^+$ ions in $ 1.00 \times 10^{-3} \, \text{M} $ aqueous solution of a weak monobasic acid having acid dissociation constant $ K_a = 4.00 \times 10^{-11} $ is $ X \times 10^{-7} \, \text{M} $. The value of $ X $ is ____. {Use: Ionic product of water $ K_w = 1.00 \times 10^{-14} $ at 25$^\circ$C
- JEE Advanced - 2025
- Ionic Equilibrium
- Consider the vectors $$ \vec{x} = \hat{i} + 2\hat{j} + 3\hat{k},\quad \vec{y} = 2\hat{i} + 3\hat{j} + \hat{k},\quad \vec{z} = 3\hat{i} + \hat{j} + 2\hat{k}. $$ For two distinct positive real numbers $ \alpha $ and $ \beta $, define $$ \vec{X} = \alpha \vec{x} + \beta \vec{y} - \vec{z},\quad \vec{Y} = \alpha \vec{y} + \beta \vec{z} - \vec{x},\quad \vec{Z} = \alpha \vec{z} + \beta \vec{x} - \vec{y}. $$ If the vectors $ \vec{X}, \vec{Y}, \vec{Z} $ lie in a plane, then the value of $ \alpha + \beta - 3 $ is ________.
- Let $ x_0 $ be the real number such that $ e^{x_0} + x_0 = 0 $. For a given real number $ \alpha $, define $$ g(x) = \frac{3xe^x + 3x - \alpha e^x - \alpha x}{3(e^x + 1)} $$ for all real numbers $ x $. Then which one of the following statements is TRUE?
- JEE Advanced - 2025
- Fundamental Theorem of Calculus
Two identical concave mirrors each of focal length $ f $ are facing each other as shown. A glass slab of thickness $ t $ and refractive index $ n_0 $ is placed equidistant from both mirrors on the principal axis. A monochromatic point source $ S $ is placed at the center of the slab. For the image to be formed on $ S $ itself, which of the following distances between the two mirrors is/are correct:
- JEE Advanced - 2025
- Ray optics and optical instruments
- Let $$ \alpha = \frac{1}{\sin 60^\circ \sin 61^\circ} + \frac{1}{\sin 62^\circ \sin 63^\circ} + \cdots + \frac{1}{\sin 118^\circ \sin 119^\circ}. $$ Then the value of $$ \left( \frac{\csc 1^\circ}{\alpha} \right)^2 $$ is \rule{1cm}{0.15mm}.
- JEE Advanced - 2025
- Some Applications of Trigonometry
JEE Advanced Notification
Concepts Used:
Stoichiometry
Stoichiometry is founded on the law of conservation of mass where the total mass of the reactants equals the total mass of the products, leading to the insight that the relations among quantities of reactants and products typically form a ratio of positive integers. This means that if the amounts of the separate reactants are known, then the amount of the product can be calculated. Conversely, if one reactant has a known quantity and the quantity of the products can be empirically determined, then the amount of the other reactants can also be calculated.
Stoichiometry helps us determine how much substance is needed or is present. Things that can be measured are;
- Reactants and Products mass
- Molecular weight
- Chemical equations
- Formulas
Stoichiometric Coefficient
The Stoichiometric coefficient of any given component is the number of molecules and/or formula units that participate in the reaction as written.
Mole Ratios
The mass of one mole of a substance in grams is called molar mass. The molar mass of one mole of a substance is numerically equal to the atomic/molecular formula mass.