Answer question 1---**SINGLE CORRECT TYPE QUESTIONS**
**Question 1:**
**Question Stem:** Above series of reactions are carried out starting with 18 g of Al and 109.5 g of HCl in first step and further 100 g of NaOH is added for step (ii) and (iii). Find out limiting reagent in each step and calculate the maximum amount of NaAlO₂ that can be produced in step (iii). (Assume reactions are taken in sequence and also that each reaction goes to 100% completion)
**Chemical Equations:**
(i) 2Al + 6HCl -> 2AlCl₃ + 3H₂
(ii) AlCl₃ + 3NaOH -> Al(OH)₃ + 3NaCl
(iii) Al(OH)₃ + NaOH -> NaAlO₂ + 2H₂O
**Options (Table Format):**
| Option | L.R. in step (I) | L.R. in step (II) | L.R. in step (III) | Moles of NaAlO₂ |
|---|---|---|---|---|
| (a) | Al | AlCl₃ | Al(OH)₃ | 0.66 |
| (b) | Al | Na(OH) | Al(OH)₃ | 0.66 |
| (c) | Al | Al | NaOH | 0.5 |
| (d) | HCl | AlCl₃ | NaOH | 0.5 |
**Question 2:**
**Question Stem:** A mixture of CH₄ and C₂H₂ was completely burnt in an excess of oxygen yielding equal volumes of CO₂ and steam, measured at the same temperature and pressure. The mole percent of CH₄ in the original mixture is
**Options:**
(a) 25%
(b) 30%
(c) 75%
(d) 50%
**Question 3:**
**Question Stem:** The strength of 10⁻² M Na₂CO₃ solution in terms of molality will be (density of solution = 1.10 g mL⁻¹). (Molecular weight of Na₂CO₃ = 106)
**Additional Data:**
density of solution = 1.10 g mL⁻¹
Molecular weight of Na₂CO₃ = 106
**Options:** (Note: Options appear to be the same as Question 2 and may be incorrect for this question based on the calculation required.)
(a) 25%
(b) 30%
(c) 75%
(d) 50%
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We have a three-step chemical reaction sequence starting with 18 grams of aluminum, 109.5 grams of hydrochloric acid, and 100 grams of sodium hydroxide. We need to find the limiting reagent in each step and calculate the maximum amount of sodium aluminate that can be produced. Let's analyze each reaction step by step.
First, we calculate the initial moles of each reactant. For aluminum: 18 grams divided by 27 grams per mole equals two-thirds mole or 0.667 moles. For hydrochloric acid: 109.5 grams divided by 36.5 grams per mole equals 3 moles. For sodium hydroxide: 100 grams divided by 40 grams per mole equals 2.5 moles. These values will be used to determine limiting reagents in each step.
Now let's analyze step one. The reaction is 2 aluminum plus 6 hydrochloric acid producing 2 aluminum chloride plus 3 hydrogen gas. The stoichiometric ratio is 1 to 3 for aluminum to HCl. We have two-thirds mole of aluminum and 3 moles of HCl. For two-thirds mole of aluminum, we need 2 moles of HCl, but we have 3 moles available. Therefore, aluminum is the limiting reagent, and we produce two-thirds mole of aluminum chloride.
For step two, aluminum chloride reacts with sodium hydroxide. We have two-thirds mole of aluminum chloride and 2.5 moles of sodium hydroxide. Aluminum chloride is limiting, producing two-thirds mole of aluminum hydroxide and leaving 0.5 moles of sodium hydroxide. In step three, aluminum hydroxide reacts with the remaining sodium hydroxide. Now sodium hydroxide becomes limiting, producing 0.5 moles of sodium aluminate. The answer is option C: aluminum, aluminum chloride, and sodium hydroxide are the limiting reagents, with 0.5 moles of sodium aluminate produced.
Let's compare our results with the given options. We found that aluminum is the limiting reagent in step one, aluminum chloride is limiting in step two, and sodium hydroxide is limiting in step three, producing 0.5 moles of sodium aluminate. Looking at the options table, option C matches our calculated limiting reagents for steps one and three, and gives the correct amount of product as 0.5 moles. Therefore, the correct answer is option C.