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Some Basic Concepts of Chemistry question

2023 · 31 Jan · Shift 2 · Q14
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Some Basic Concepts of Chemistry question

2023 · 31 Jan · Shift 2 · Q14

JEE MainChemistrySome Basic Concepts of ChemistryNumerical+4 / −1
Assume carbon burns according to the following equation : 2C(s)+O2( g)→2CO(g)2 \mathrm{C}_{(\mathrm{s})}+\mathrm{O}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{CO}(\mathrm{g})2C(s)​+O2( g)​→2CO(g) when 12 g12 \mathrm{~g}12 g carbon is burnt in 48 g48 \mathrm{~g}48 g of oxygen, the volume of carbon monoxide produced is ‾\underline{\hspace{2cm}}​×10−1 L\times 10^{-1} \mathrm{~L}×10−1 L at STP [nearest integer] [Given: Assume CO\mathrm{CO}CO as ideal gas, Mass of C\mathrm{C}C is 12 g mol−112 \mathrm{~g} \mathrm{~mol}^{-1}12 g mol−1, Mass of O\mathrm{O}O is 16 g mol−116 \mathrm{~g} \mathrm{~mol}^{-1}16 g mol−1 and molar volume of an ideal gas at STP is 22.7 L mol−122.7 \mathrm{~L} \mathrm{~mol}^{-1}22.7 L mol−1 ]
Numerical answer
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Correct answer: 227

  1. Write the balanced reaction

2C(s)+O2(g)→2CO(g)2C_{(s)} + O_{2(g)} \rightarrow 2CO_{(g)}2C(s)​+O2(g)​→2CO(g)​

  1. Calculate moles of reactants

  • Moles of carbon: n(C)=1212=1 moln(C)=\frac{12}{12}=1\text{ mol}n(C)=1212​=1 mol

  • Moles of oxygen: n(O2)=4832=1.5 moln(O_2)=\frac{48}{32}=1.5\text{ mol}n(O2​)=3248​=1.5 mol

  1. Find the limiting reagent

From the reaction, 2 mol C react with 1 mol O22\text{ mol } C \text{ react with } 1\text{ mol } O_22 mol C react with 1 mol O2​

So, for 111 mol CCC, required O2O_2O2​ is: 12=0.5 mol\frac{1}{2}=0.5\text{ mol}21​=0.5 mol

Available O2=1.5O_2 = 1.5O2​=1.5 mol, so oxygen is in excess and carbon is the limiting reagent.

  1. Calculate moles of CO formed

From the equation, 2C→2CO2C \rightarrow 2CO2C→2CO

Thus, 1 mol C→1 mol CO1\text{ mol } C \rightarrow 1\text{ mol } CO1 mol C→1 mol CO

So moles of CO produced: n(CO)=1 moln(CO)=1\text{ mol}n(CO)=1 mol

  1. Calculate volume of CO at STP

Given molar volume at STP = 22.7 L mol−122.7\,L\,mol^{-1}22.7Lmol−1

Hence, V(CO)=1×22.7=22.7 LV(CO)=1\times 22.7=22.7\,LV(CO)=1×22.7=22.7L

  1. Convert into the asked form

Question asks for: ‾×10−1 L\underline{\hspace{2cm}}\times 10^{-1}\,L​×10−1L

Now, 22.7 L=227×10−1 L22.7\,L = 227 \times 10^{-1}\,L22.7L=227×10−1L

Therefore, the required nearest integer is: 227\boxed{227}227​

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