### Related Formula textMoles (n) = fractextMass in gramstextMolar Mass$\text{Moles } (n) = \frac{\text{Mass in grams}}{\text{Molar Mass}}$ textVolume of water (V) = fractextMass of watertextDensity of water$\text{Volume of water } (V) = \frac{\text{Mass of water}}{\text{Density of water}}$ ### Core Logic First, calculate initial molar quantities for both reactants: - Molar mass of textC_4textH_10 = 4(12) + 10(1) = 58text g/mol$\text{C}_4\text{H}_{10} = 4(12) + 10(1) = 58\text{ g/mol}$ - Initial Moles of butane = frac174.0 times 10^3text g58text g/mol = 3000text mol = 3 times 10^3text mol $= \frac{174.0 \times 10^3\text{ g}}{58\text{ g/mol}} = 3000\text{ mol} = 3 \times 10^3\text{ mol} $ - Molar mass of textO_2 = 32text g/mol$\text{O}_2 = 32\text{ g/mol}$ - Initial Moles of oxygen = frac320.0 times 10^3text g32text g/mol = 10000text mol = 10 times 10^3text mol $= \frac{320.0 \times 10^3\text{ g}}{32\text{ g/mol}} = 10000\text{ mol} = 10 \times 10^3\text{ mol} $ ### Step 1: Identify Limiting Reagent Let's test the stoichiometric requirements via calculation ratios: - Ratio for textC_4textH_10 = frac30001 = 3000$\text{C}_4\text{H}_{10} = \frac{3000}{1} = 3000$ - Ratio for textO_2 = frac1000013/2 = 1538.46$\text{O}_2 = \frac{10000}{13/2} = 1538.46$ Since the ratio for textO_2$\text{O}_2$ is lower, oxygen behaves as the **limiting reagent** and commands the output steps. ### Step 2: Compute Water Yield Using stoichiometric proportions defined by the balanced reaction field: textMoles of textH2textO = 5 times fractextMoles of textO213/2 = 5 times frac213 times 10000 = frac10000013text mol $\text{Moles of } \text{H}2\text{O} = 5 \times \frac{\text{Moles of } \text{O}2}{13/2} = 5 \times \frac{2}{13} \times 10000 = \frac{100000}{13}\text{ mol} $ Convert moles to mass (MtextH2textO = 18text g/mol$M{\text{H}2\text{O}} = 18\text{ g/mol}$): textMass of water = frac10000013 times 18 = 138461.5text g approx 138.46text kg $\text{Mass of water} = \frac{100000}{13} \times 18 = 138461.5\text{ g} \approx 138.46\text{ kg} $ Since density = 1text g/mL = 1text kg/L$= 1\text{ g/mL} = 1\text{ kg/L}$, the net volume is exactly: Vtextwater = 138.46text Litres approx 138text Litres $V{\text{water}} = 138.46\text{ Litres} \approx 138\text{ Litres} $ ### Pattern Recognition Limiting reagent shortcut: Always check ratios (moles / stoichiometric coefficient) right away. Do not spend time calculating theoretical products based on butane before establishing whether oxygen runs out first. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry

### Core Logic Let's find the empirical formula by calculating the relative molar ratios of each element: * Carbon (C): textMoles = frac54.212 = 4.516$\text{Moles} = \frac{54.2}{12} = 4.516$ * Hydrogen (H): textMoles = frac9.21 = 9.200$\text{Moles} = \frac{9.2}{1} = 9.200$ * Oxygen (O): textMoles = frac36.616 = 2.287$\text{Moles} = \frac{36.6}{16} = 2.287$ Next, divide each value by the smallest number of moles (2.287$2.287$) to get the simplest whole-number ratio: * C: frac4.5162.287 approx 1.97 approx 2$\frac{4.516}{2.287} \approx 1.97 \approx 2$ * H: frac9.2002.287 approx 4.02 approx 4$\frac{9.200}{2.287} \approx 4.02 \approx 4$ * O: frac2.2872.287 = 1$\frac{2.287}{2.287} = 1$ Thus, the Empirical Formula is mathrmC_2mathrmH_4mathrmO$\mathrm{C}_{2}\mathrm{H}_{4}\mathrm{O}$. ### Step 2: Molecular Formula Determination 1. Calculate the empirical formula mass: textEmpirical Mass = (2 cdot 12) + (4 cdot 1) + 16 = 24 + 4 + 16 = 44text g/mol$\text{Empirical Mass} = (2 \cdot 12) + (4 \cdot 1) + 16 = 24 + 4 + 16 = 44\text{ g/mol}$ 2. Find the multiplier (n$n$): n = fractextMolar MasstextEmpirical Mass = frac13244 = 3$n = \frac{\text{Molar Mass}}{\text{Empirical Mass}} = \frac{132}{44} = 3$ 3. Determine the molecular formula: textMolecular Formula = (mathrmC_2mathrmH_4mathrmO)_3 = mathrmC_6mathrmH_12mathrmO_3$\text{Molecular Formula} = (\mathrm{C}_{2}\mathrm{H}_{4}\mathrm{O})_{3} = \mathrm{C}_{6}\mathrm{H}_{12}\mathrm{O}_{3}$ This matches Option (3). ### Pattern Recognition Always determine the empirical formula ratio first. Then, compare the empirical mass with the given molar mass to find the whole-number multiplier (n$n$). Multiplication gives the definitive molecular formula. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry

### Related Formula textMoles (n) = fractextMass in gramstextMolar Mass$\text{Moles } (n) = \frac{\text{Mass in grams}}{\text{Molar Mass}}$ ### Core Logic First, calculate the input moles for each reactant: - Moles of Fe_3O_4 = frac2.32 times 10^3 times 10^3text g232text g mol^-1 = 10,000text moles$Fe_{3}O_{4} = \frac{2.32 \times 10^3 \times 10^3\text{ g}}{232\text{ g mol}^{-1}} = 10,000\text{ moles}$ - Moles of CO = frac2.8 times 10^2 times 10^3text g28text g mol^-1 = 10,000text moles$CO = \frac{2.8 \times 10^2 \times 10^3\text{ g}}{28\text{ g mol}^{-1}} = 10,000\text{ moles}$ Next, identify the limiting reagent by comparing the available moles to the stoichiometric coefficients: - For Fe_3O_4$Fe_3O_4$: frac100001 = 10000$\frac{10000}{1} = 10000$ - For CO$CO$: frac100004 = 2500$\frac{10000}{4} = 2500$ Since 2500 < 10000$2500 < 10000$, **carbon monoxide (CO$CO$) is the limiting reagent**. Now, determine the production yield of iron based on the limiting reagent (CO$CO$): textMoles of Fe text produced = frac34 times n(CO) = frac34 times 10000 = 7500text moles$\text{Moles of } Fe \text{ produced} = \frac{3}{4} \times n(CO) = \frac{3}{4} \times 10000 = 7500\text{ moles}$ Convert these moles into kilograms to find the final mass (x$x$): textMass of Fe = frac7500 times 56text g/mol1000text g/kg = 420text kg$\text{Mass of } Fe = \frac{7500 \times 56\text{ g/mol}}{1000\text{ g/kg}} = 420\text{ kg}$ ### Pattern Recognition Always identify the limiting reagent first by normalizing the mole quantities with their respective stoichiometric coefficients before calculating product yields. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry

### Related Formula Mass percentage definition: \%text w/w = fractextMass of solutetextMass of solution times 100$\%\text{ w/w} = \frac{\text{Mass of solute}}{\text{Mass of solution}} \times 100$ Density conversion equation: textVolume of solution = fractextMass of solutiontextDensity of solution$\text{Volume of solution} = \frac{\text{Mass of solution}}{\text{Density of solution}}$ ### Core Logic A value of 75\%text w/w HNO_3$75\%\text{ w/w HNO}_3$ implies that 75mathrm\ g$75\mathrm{\ g}$ of pure textHNO_3$\text{HNO}_3$ is present in 100mathrm\ g$100\mathrm{\ g}$ of solution. We need to find the volume that provides exactly 30mathrm\ g$30\mathrm{\ g}$ of pure acid solute. ### Step 1: Calculate Solution Mass and Volume Mass of solution needed for 30mathrm\ g$30\mathrm{\ g}$ solute: textMass = frac10075 times 30 = 40mathrm\ g$\text{Mass} = \frac{100}{75} \times 30 = 40\mathrm{\ g}$ Converting mass to volume using solution density (1.25mathrm\ g/mL$1.25\mathrm{\ g/mL}$): textVolume = frac40mathrm\ g1.25mathrm\ g/mL = 32mathrm\ mL$\text{Volume} = \frac{40\mathrm{\ g}}{1.25\mathrm{\ g/mL}} = 32\mathrm{\ mL}$ ### Pattern Recognition Break concentration steps down clearly: textMass of solute rightarrow textMass of solution rightarrow textVolume of solution$\text{Mass of solute} \rightarrow \text{Mass of solution} \rightarrow \text{Volume of solution}$. Combining operations: textVolume = fractextMass solute\% times frac100textdensity = frac3075 times frac1001.25 = 0.4 times 80 = 32$\text{Volume} = \frac{\text{Mass solute}}{\%} \times \frac{100}{\text{density}} = \frac{30}{75} \times \frac{100}{1.25} = 0.4 \times 80 = 32$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry

### Related Formula T_mathrmK = T_^circmathrmC + 273.15$T_{\mathrm{K}} = T_{^{\circ}\mathrm{C}} + 273.15$ Absolute zero (0text K$0\text{ K}$) represents the lowest theoretical temperature limit. ### Core Logic Analyzing each statement based on foundational definitions : * (A) & (B) Incorrect: Mass is the actual matter present; weight is the gravitational force exerted on that mass. These definitions are reversed in the statements. * (C) Correct: Volume correctly defines the space occupied by a substance . * (D) Correct: Celsius values can be negative, whereas Kelvin scale strictly defaults to absolute zero (0text K$0\text{ K}$) as minimum . * (E) Correct: Precision measures how close experimental trials lie relative to each other . Therefore, statements (C), (D), and (E) are correct. ### Pattern Recognition Absolute temperature scale (Kelvin) can never possess real negative values because 0text K$0\text{ K}$ represents complete cessation of molecular motion. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry