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Some Basic Concepts of Chemistry question

2023 · 11 Apr · Shift 2 · Q22
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Some Basic Concepts of Chemistry question

2023 · 11 Apr · Shift 2 · Q22

JEE MainChemistrySome Basic Concepts of ChemistryNumerical+4 / −1
The volume of hydrogen liberated at STP by treating 2.4 g2.4 \mathrm{~g}2.4 g of magnesium with excess of hydrochloric acid is ‾\underline{\hspace{2cm}}​× 10−2 L\times ~10^{-2} \mathrm{~L}× 10−2 L Given : Molar volume of gas is 22.4 L22.4 \mathrm{~L}22.4 L at STP. Molar mass of magnesium is 24 g mol−124 \mathrm{~g} \mathrm{~mol}^{-1}24 g mol−1
Numerical answer
View written solutionFree

Correct answer: 224

  1. Write the reaction

Mg+2HCl→MgCl2+H2\mathrm{Mg + 2HCl \rightarrow MgCl_2 + H_2}Mg+2HCl→MgCl2​+H2​

From the balanced equation,

  • 111 mole of Mg\mathrm{Mg}Mg gives 111 mole of H2\mathrm{H_2}H2​.
  1. Calculate moles of magnesium

Given mass of magnesium =2.4 g= 2.4\,\mathrm{g}=2.4g

Molar mass of magnesium =24 g mol−1= 24\,\mathrm{g\,mol^{-1}}=24gmol−1

n(Mg)=2.424=0.1 moln(\mathrm{Mg}) = \frac{2.4}{24} = 0.1\,\mathrm{mol}n(Mg)=242.4​=0.1mol

So,

n(H2)=0.1 moln(\mathrm{H_2}) = 0.1\,\mathrm{mol}n(H2​)=0.1mol

  1. Calculate volume of hydrogen at STP

At STP, molar volume =22.4 L= 22.4\,\mathrm{L}=22.4L

V(H2)=0.1×22.4=2.24 LV(\mathrm{H_2}) = 0.1 \times 22.4 = 2.24\,\mathrm{L}V(H2​)=0.1×22.4=2.24L

  1. Express in the asked form

We need

V=‾×10−2 LV = \underline{\hspace{2cm}} \times 10^{-2}\,\mathrm{L}V=​×10−2L

Since,

2.24 L=224×10−2 L2.24\,\mathrm{L} = 224 \times 10^{-2}\,\mathrm{L}2.24L=224×10−2L

So the required integer is:

224\boxed{224}224​

  1. Comparison with stored answer

Stored correct answer = 224224224

Our derived answer also is 224224224, so they agree.

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