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Some Basic Concepts of Chemistry question

2019 · 10 Apr · Shift 2 · Q9
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Some Basic Concepts of Chemistry question

2019 · 10 Apr · Shift 2 · Q9

JEE MainChemistrySome Basic Concepts of ChemistryMCQ+4 / −1
The minimum amount of O2O_2O2​(g) consumed per gram of reactant is for the reaction : (Given atomic mass : Fe = 56, O = 16, Mg = 24, P = 31, C = 12, H = 1)
  1. A
    4FeFeFe(s) + 3O2O_2O2​(g) →\to→ 2Fe2O3Fe_2O_3Fe2​O3​(s)
  2. B
    P4P_4P4​(s) + 5O2O_2O2​(g) →\to→ P4O10P_4O_{10}P4​O10​(s)
  3. C
    C3H8C_3H_8C3​H8​(g) + 5O2O_2O2​(g) →\to→ 3CO2CO_2CO2​(g) + 4H2OH_2OH2​O(l)
  4. D
    2MgMgMg(s) + O2O_2O2​(g) →\to→ 2MgOMgOMgO(s)
View written solutionFree

Correct answer: A

We need to find which reaction consumes the minimum mass of O2O_2O2​ per gram of reactant.

1. Strategy

For each reaction, compute:

mass of O2 consumedmass of reactant consumed\frac{\text{mass of } O_2 \text{ consumed}}{\text{mass of reactant consumed}}mass of reactant consumedmass of O2​ consumed​

Here, “reactant” means the non-O2O_2O2​ reactant given on the left side.

The reaction with the smallest value is the correct option.

2. Evaluate each option

Option A

4Fe+3O2→2Fe2O34Fe + 3O_2 \to 2Fe_2O_34Fe+3O2​→2Fe2​O3​

Mass of 4Fe4Fe4Fe:

4×56=2244 \times 56 = 2244×56=224

Mass of 3O23O_23O2​:

3×32=963 \times 32 = 963×32=96

So, O2O_2O2​ consumed per gram of Fe is:

96224=37≈0.429\frac{96}{224} = \frac{3}{7} \approx 0.42922496​=73​≈0.429

Option B

P4+5O2→P4O10P_4 + 5O_2 \to P_4O_{10}P4​+5O2​→P4​O10​

Mass of P4P_4P4​:

4×31=1244 \times 31 = 1244×31=124

Mass of 5O25O_25O2​:

5×32=1605 \times 32 = 1605×32=160

So, O2O_2O2​ consumed per gram of P4P_4P4​ is:

160124=4031≈1.290\frac{160}{124} = \frac{40}{31} \approx 1.290124160​=3140​≈1.290

Option C

C3H8+5O2→3CO2+4H2OC_3H_8 + 5O_2 \to 3CO_2 + 4H_2OC3​H8​+5O2​→3CO2​+4H2​O

Mass of C3H8C_3H_8C3​H8​:

3(12)+8(1)=36+8=443(12) + 8(1) = 36 + 8 = 443(12)+8(1)=36+8=44

Mass of 5O25O_25O2​:

5×32=1605 \times 32 = 1605×32=160

So, O2O_2O2​ consumed per gram of C3H8C_3H_8C3​H8​ is:

16044=4011≈3.636\frac{160}{44} = \frac{40}{11} \approx 3.63644160​=1140​≈3.636

Option D

2Mg+O2→2MgO2Mg + O_2 \to 2MgO2Mg+O2​→2MgO

Mass of 2Mg2Mg2Mg:

2×24=482 \times 24 = 482×24=48

Mass of O2O_2O2​:

323232

So, O2O_2O2​ consumed per gram of Mg is:

3248=23≈0.667\frac{32}{48} = \frac{2}{3} \approx 0.6674832​=32​≈0.667

3. Compare all values

  • A: 96224=0.429\displaystyle \frac{96}{224} = 0.42922496​=0.429
  • B: 160124=1.290\displaystyle \frac{160}{124} = 1.290124160​=1.290
  • C: 16044=3.636\displaystyle \frac{160}{44} = 3.63644160​=3.636
  • D: 3248=0.667\displaystyle \frac{32}{48} = 0.6674832​=0.667

The minimum value is for Option A.

4. Final Answer

A\boxed{A}A​

The minimum amount of O2O_2O2​ consumed per gram of reactant is in:

4Fe+3O2→2Fe2O34Fe + 3O_2 \to 2Fe_2O_34Fe+3O2​→2Fe2​O3​

5. Comparison with stored correct answer

Stored correct answer: A

My derived answer: A

So, the answer agrees with the stored correct answer.

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