Let's carefully analyze and solve the problem step by step. --- ### Given Data: - 1 gram atom (equivalent to 1 mole) of metal reacts completely with water. - The hydroxide solution is made up to **1 liter**. - **100 mL** of this solution requires **11 mL** of **0.9091 N HCl** for neutralization. - The chloride metal contains **13.7% Cl** by weight. - Find the volume of **H₂ gas** liberated at **STP**. --- ### Step 1: Determine the amount of chloride metal in 1 liter of solution. Since 1 liter of solution contains the hydroxide formed from 1 mole of metal, and the metal has reacted completely, the total amount of chloride in the solution corresponds to the chloride formed from the metal. --- ### Step 2: Find the amount of chloride in 100 mL of solution. Given: - 100 mL of solution requires 11 mL of 0.9091 N HCl. The normality (N) relates to the equivalent amount of HCl needed for neutralization. Number of equivalents of HCl in 11 mL: \[ \text{Equivalents} = N \times \text{volume (L)} = 0.9091 \times 0.011 \text{ L} = 0.0100 \text{ equivalents} \] Since HCl is a monoprotic acid, 1 equivalent of HCl neutralizes 1 equivalent of base. Thus, **equivalents of hydroxide in 100 mL of solution = 0.0100 equivalents**. --- ### Step 3: Determine the total equivalents in 1 liter. Since 100 mL contains 0.0100 equivalents, 1 liter (1000 mL) contains: \[ \frac{1\, \text{L}}{0.1\, \text{L}} \times 0.0100\, \text{equiv} = 10 \times 0.0100 = 0.10\, \text{equiv} \] **Therefore, total hydroxide equivalents in 1 liter = 0.10 equivalents.** --- ### Step 4: Find the molar amount of chloride in the solution. Since the hydroxide came from the metal chloride (assuming MCl), the equivalents of hydroxide correspond to the equivalents of chloride in the original chloride compound. Given: - The chloride metal contains **13.7% Cl** by weight. - The total chloride equivalents are 0.10 equivalents in 1 liter. --- ### Step 5: Calculate the moles of chloride. - **Number of equivalents of chloride** is equal to the total equivalents because 1 mol of MCl releases 1 mol of Cl⁻ (assuming monovalent chloride). - Moles of Cl in the solution: \[ \text{Moles of Cl} = \text{equivalents} = 0.10\, \text{moles} \] --- ### Step 6: Find the mass of chloride in the solution. Mass of Cl in the solution: \[ \text{Mass of Cl} = \text{moles} \times \text{atomic mass of Cl} = 0.10 \times 35.45\, \text{g/mol} = 3.545\, \text{g} \] ### Step 7: Determine the total mass of chloride metal present. Since chloride is 13.7% of the total metal by weight: \[ \text{Mass of metal} = \frac{\text{Mass of Cl}}{0.137} = \frac{3.545\, \text{g}}{0.137} \approx 25.89\, \text{g} \] --- ### Step 8: Find the equivalent weight of the metal. - 1 mole of metal reacts with water to produce hydroxide and hydrogen gas. - Given 1 mole of metal corresponds to 25.89 g (from previous calculation). - The metal's valency (number of electrons involved per atom) depends on the metal's properties, but since it forms a hydroxide with the formula MOH, it's likely monovalent (e.g., Na, K). --- ### Step 9: Find the volume of H₂ gas at STP. - The reaction of metal with water: \[ \text{M} + \text{H}_2\text{O} \rightarrow \text{MOH} + \text{H}_2 \uparrow \] - 1 mole of metal produces 1 mole of H₂ gas. - Since 1 mole of metal weighs 25.89 g, and in total, 1 mole (from the original 1 gram atom) of metal reacted, producing: \[ \text{moles of H}_2 = 1\, \text{mol} \] - **But wait**, earlier, we calculated the total chloride in the solution based on the chloride content, which corresponds to 0.10 equivalents (or 0.10 mol of Cl). Since the chloride corresponds to the metal, and the initial metal was 1 mole (gram atom), but the total chloride in the solution is only 0.10 mol, indicating the amount of metal involved is 0.10 mol. **Therefore, the actual moles of metal reacting: 0.10 mol.** --- ### Step 10: Volume of H₂ gas liberated at STP. - 1 mol of H₂ gas occupies 22.4 L at STP. - For 0.10 mol: \[ V_{H_2} = 0.10 \times 22.4\, \text{L} = 2.24\, \text{L} \] --- ### **Final answer:** \[ \boxed{ \text{Volume of } H_2 \text{ gas} \approx \mathbf{2.24\, \text{liters}} } \] --- ## **Summary:** - The amount of chloride in the solution corresponds to 0.10 mol. - The moles of metal reacting are 0.10 mol. - At STP, 0.10 mol of H₂ gas is liberated, which equals **2.24 liters**. **Therefore, the volume of hydrogen gas evolved at STP is approximately 2.24 liters.**

Thank you for correcting the typo. **Step 2**: The correct volume is **110 mL** (not 11 mL) of 0.9091 N HCl. Let's recalculate using this corrected value. --- ## **Step 2 (Revised): Calculate the total equivalents of base in 100 mL of solution** Given: - 100 mL sample required **110 mL** of **0.9091 N HCl**. Equivalents of HCl used: \[ \text{Equivalents} = N \times V (\text{in L}) = 0.9091 \times 0.110 = 0.100 \text{ equiv} \] So, **100 mL of the solution contains 0.100 equivalents of base**. --- ## **Step 3 (Revised): Calculate the total equivalents in 1 liter** Since 100 mL contains 0.100 equivalents, 1 L (1000 mL) contains: \[ \frac{1000}{100} \times 0.100 = 10 \times 0.100 = 1.00 \text{ equivalents} \] --- ## **Step 4 (Revised): Relate equivalents to moles of metal (hydroxide) formed** If the metal forms MOH (monovalent, like Na), 1 equivalent = 1 mole, so there are **1 mole of MOH** in 1 L. Thus, **1 mole of metal reacted**, which matches the data (1 gram atom of metal). --- ## **Step 5: Volume of H₂ liberated at STP** Reaction: \[ M + H_2O \to MOH + \frac{1}{2} H_2 \] So, **1 mole of metal gives 0.5 mole H₂**. Volume at STP: \[ 0.5 \text{ mol} \times 22.4 \text{ L/mol} = 11.2 \text{ L} \] --- ## **Step 6: Cross-Check with Cl Content** - The metal chloride contains 13.7% Cl by weight. - 1 mole of MOH formed means originally 1 mole of MCl was present (if the metal started as MCl). - 1 mole of MCl contains 1 mole Cl = 35.5 g. If the molar mass of MCl is \( x \): \[ \frac{35.5}{x} = 0.137 \implies x = \frac{35.5}{0.137} \approx 259 \] So molar mass of M = 259 - 35.5 ≈ 223.5 g/mol --- ## **Final Answer:** \[ \boxed{11.2~\text{liters}} \] **The volume of H₂ gas liberated at STP is \( \mathbf{11.2~\text{liters}} \).**