textA + 2textB rightarrow textAB_2$\text{A} + 2\text{B} \rightarrow \text{AB}_2$ 36.0text g$36.0\text{ g}$ of 'A' (Molar mass: 60text g mol^-1$60\text{ g mol}^{-1}$) and 56.0text g$56.0\text{ g}$ of 'B' (Molar mass: 80text g mol^-1$80\text{ g mol}^{-1}$) are allowed to react. Which of the following statements are correct? (A) 'A' is the limiting reagent (B) 77.0text g$77.0\text{ g}$ of textAB_2$\text{AB}_2$ is formed (C) Molar mass of textAB_2$\text{AB}_2$ is 140text g mol^-1$140\text{ g mol}^{-1}$ (D) 15.0text g$15.0\text{ g}$ of A is left unreacted after the completion of reaction. Choose the correct answer from the options given below:

### Related Formula mathrmC_mathrmxmathrmH_mathrmy(g) + left(mathrmx + fracmathrmy4right)mathrmO_2(g) longrightarrow mathrmxCO_2(g) + fracmathrmy2mathrmH_2mathrmO_(ell)$$\mathrm{C}_\mathrm{x}\mathrm{H}_\mathrm{y(g)} + \left(\mathrm{x} + \frac{\mathrm{y}}{4}\right)\mathrm{O}_{2(g)} \longrightarrow \mathrm{xCO}_{2(g)} + \frac{\mathrm{y}}{2}\mathrm{H}_2\mathrm{O}_{(\ell)}$$ ### Core Logic Let the volume of hydrocarbon be V = 80$V = 80$ mL. Initial volume of O_2 = 264$O_2 = 264$ mL. At 273 K, H_2O$H_2O$ is liquid, so its volume is neglected. Volume of CO_2$CO_2$ formed = 80x$80x$ mL. Volume of O_2$O_2$ used = 80left(x + fracy4right)$80\left(x + \frac{y}{4}\right)$ mL. Unreacted O_2$O_2$ = 264 - 80left(x + fracy4right)$264 - 80\left(x + \frac{y}{4}\right)$ mL. Total residual volume = V_CO_2 + V_unreacted \ O_2 = 224$V_{CO_2} + V_{unreacted \ O_2} = 224$ mL. 80x + 264 - 80left(x + fracy4right) = 224$$80x + 264 - 80\left(x + \frac{y}{4}\right) = 224$$ 264 - frac80y4 = 224$$264 - \frac{80y}{4} = 224$$ 40 = 20y implies y = 2$$40 = 20y \implies y = 2$$ After treatment with KOH, CO_2$CO_2$ is absorbed. The remaining volume is unreacted O_2$O_2$, which is 64 mL. 264 - 80left(x + fracy4right) = 64$$264 - 80\left(x + \frac{y}{4}\right) = 64$$ Substitute y = 2$y = 2$: 264 - 80left(x + frac12right) = 64$$264 - 80\left(x + \frac{1}{2}\right) = 64$$ 264 - 80x - 40 = 64$$264 - 80x - 40 = 64$$ 224 - 80x = 64$224 - 80x = 64$ 80x = 160 implies x = 2$$80x = 160 \implies x = 2$$ The hydrocarbon is mathrmC_2mathrmH_2$\mathrm{C}_2\mathrm{H}_2$. ### Pattern Recognition Volume decrease by KOH indicates the volume of CO_2$CO_2$ produced. V_CO_2 = 224 - 64 = 160$V_{CO_2} = 224 - 64 = 160$ mL. V_HC = 80$V_{HC} = 80$ mL. So x = frac16080 = 2$x = \frac{160}{80} = 2$. Total volume reduction = 264 - 64 = 200 mL (O_2$O_2$ consumed). O_2$O_2$ consumed = 80(x + y/4) = 200 implies 2 + y/4 = 2.5 implies y/4 = 0.5 implies y = 2$80(x + y/4) = 200 \implies 2 + y/4 = 2.5 \implies y/4 = 0.5 \implies y = 2$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry

### Related Formula mathrmCa_(mathrms) + 2mathrmHCl_(mathrmg) longrightarrow mathrmCaCl_2(mathrms) + mathrmH_2(mathrmg)$$\mathrm{Ca}_{(\mathrm{s})} + 2\mathrm{HCl}_{(\mathrm{g})} \longrightarrow \mathrm{CaCl}_{2(\mathrm{s})} + \mathrm{H}_{2(\mathrm{g})}$$ ### Core Logic Number of moles of Calcium (n_Ca$n_{Ca}$) = fractextGiven masstextMolar mass = frac14.040 = 0.35$\frac{\text{Given mass}}{\text{Molar mass}} = \frac{14.0}{40} = 0.35$ mol. Since HCl is in excess, Calcium is the limiting reagent. From stoichiometry, 1 mole of Ca produces 1 mole of H_2$H_2$ gas and 1 mole of CaCl_2$CaCl_2$. Moles of H_2$H_2$ evolved = 0.35 mol. Volume of H_2$H_2$ at STP (1 atm, 273 K) = 0.35 times 22.4$0.35 \times 22.4$ L = 7.84$7.84$ L. Mass of CaCl_2$CaCl_2$ produced = 0.35 times textMolar mass of CaCl_2$0.35 \times \text{Molar mass of } CaCl_2$ = 0.35 times (40 + 71) = 0.35 times 111 = 38.85$0.35 \times (40 + 71) = 0.35 \times 111 = 38.85$ g. Option (3) states 33.3 g of CaCl_2$CaCl_2$ is produced, which is incorrect. ### Step 1: Final Conclusion Statement (3) is incorrect. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry

### Related Formula \% text of textP = fractextMoles of textMg_2textP_2textO_7 times 2 times 31textMass of compound times 100$$\% \text{ of } \text{P} = \frac{\text{Moles of } \text{Mg}_2\text{P}_2\text{O}_7 \times 2 \times 31}{\text{Mass of compound}} \times 100$$ ### Core Logic Molar mass of textMg_2textP_2textO_7 = 2(24) + 2(31) + 7(16) = 48 + 62 + 112 = 222 text g/mol$\text{Mg}_2\text{P}_2\text{O}_7 = 2(24) + 2(31) + 7(16) = 48 + 62 + 112 = 222 \text{ g/mol}$. textPercentage of P = fracleft(frac1.79222 times 2 times 31right)1 times 100 = 49.99\% approx 50\%$$\text{Percentage of P} = \frac{\left(\frac{1.79}{222} \times 2 \times 31\right)}{1} \times 100 = 49.99\% \approx 50\%$$ ### Step 1: Final Calculation Rounding to the nearest integer gives 50. ### Pattern Recognition Sees: Gravimetric analysis involving magnesium pyrophosphate. Trap: Forgetting the factor of 2 for phosphorus atoms in textMg_2textP_2textO_7$\text{Mg}_2\text{P}_2\text{O}_7$. ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry

### Related Formula textVolume of gas at STP = textMoles times 22.4 text L$$\text{Volume of gas at STP} = \text{Moles} \times 22.4 \text{ L}$$ ### Core Logic From the balanced chemical equation: 2Al(s) + 6HCl(aq) rightarrow 2Al^3+(aq) + 6Cl^-(aq) + 3H_2(g)$$2Al(s) + 6HCl(aq) \rightarrow 2Al^{3+}(aq) + 6Cl^{-}(aq) + 3H_{2}(g)$$ 6 moles of HCl produce 3 moles of H_2$H_2$. Therefore, 1 mole of HCl produces frac36 = 0.5$\frac{3}{6} = 0.5$ moles of H_2$H_2$. Volume of H_2$H_2$ produced at STP for 1 mole of HCl: V = 0.5 times 22.4text L = 11.2text L$$V = 0.5 \times 22.4\text{ L} = 11.2\text{ L}$$ ### Step 1: Check other options Option (2): 2 moles of Al produce 3 moles of H_2$H_2$. 1 mole Al produces 1.5 moles H_2 = 1.5 times 22.4 = 33.6text L$H_2 = 1.5 \times 22.4 = 33.6\text{ L}$ (Incorrect). Option (4) is incorrect because volume depends on temperature and pressure. ### Pattern Recognition Standard stoichiometry. Directly map the mole ratio from the balanced equation to molar volume at standard conditions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry