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Current Question (ID: 7398)

Question:

$\text{The moles of lead (II) chloride that will be formed from a reaction between 6.5 g of PbO and 3.2 g of HCl are:}$

Options:

1. $0.044$
2. $0.333$
3. $0.011$
4. $0.029$

Solution:

$\text{Hint: Equation based mole concept related questions and limiting reagent.}$\n\n$\text{Step 1: First, we need to write and balance the chemical equation for the reaction.}$\n\n$\text{PbO} + 2\text{HCl} \rightarrow \text{PbCl}_2 + \text{H}_2\text{O}$\n\n$\text{Step 2: Calculate the molecular weights of the reactants and products}$\n\n$\text{Molecular weight of PbO} = 207.2 + 16 = 223.2 \text{ g/mol}$\n\n$\text{Molecular weight of HCl} = 35.5 + 1 = 36.5 \text{ g/mol}$\n\n$\text{Molecular weight of } \text{PbCl}_2 = 207.2 + 2(35.5) = 278.2 \text{ g/mol}$\n\n$\text{Step 3: Calculate the number of moles of each reactant}$\n\n$\text{Moles of PbO} = \frac{6.5 \text{ g}}{223.2 \text{ g/mol}} = 0.0291 \text{ mol}$\n\n$\text{Moles of HCl} = \frac{3.2 \text{ g}}{36.5 \text{ g/mol}} = 0.0877 \text{ mol}$\n\n$\text{Step 4: Determine the limiting reagent}$\n\n$\text{From the balanced equation, 1 mole of PbO reacts with 2 moles of HCl}$\n\n$\text{For 0.0291 moles of PbO, we need } 2 \times 0.0291 = 0.0582 \text{ moles of HCl}$\n\n$\text{Since we have 0.0877 moles of HCl, which is more than required, PbO is the limiting reagent.}$\n\n$\text{Step 5: Calculate the moles of } \text{PbCl}_2 \text{ formed}$\n\n$\text{According to the balanced equation, 1 mole of PbO produces 1 mole of } \text{PbCl}_2$\n\n$\text{Therefore, 0.0291 moles of PbO will produce 0.0291 moles of } \text{PbCl}_2$\n\n$\text{Alternatively, we can calculate this using the mass relationship:}$\n\n$\text{Since 223.2 g PbO gives 278.2 g } \text{PbCl}_2$\n\n$6.5 \text{ g PbO will give } \frac{278.2}{223.2} \times 6.5 \text{ g of } \text{PbCl}_2 = 8.11 \text{ g}$\n\n$\text{Converting to moles: } \frac{8.11 \text{ g}}{278.2 \text{ g/mol}} = 0.029 \text{ mol of } \text{PbCl}_2$\n\n$\text{Therefore, the moles of lead (II) chloride formed will be 0.029 mol.}$

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Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}