JEE Challenger
More from Some Basic Concepts of Chemistry
ChemMedAdv 2020 • Paper 1

Volume of Hydrogen Gas Produced from Reaction of Aluminium and Sulfuric Acid

Aluminium reacts with sulfuric acid to produce aluminium sulfate and hydrogen gas. What is the total volume of hydrogen gas in liters (L) generated at 300 K300\text{ K} and 1.0 atm1.0\text{ atm} pressure when 5.4 g5.4\text{ g} of aluminium is combined with 50.0 mL50.0\text{ mL} of 5.0 M5.0\text{ M} sulfuric acid solution for the reaction? (Use atomic mass of aluminium =27.0 g mol−1= 27.0\text{ g mol}^{-1}, R=0.082 atm L mol−1 K−1R = 0.082\text{ atm L mol}^{-1}\text{ K}^{-1})

Official Numerical Answer6.15

Topics & Concepts

Some Basic Concepts of ChemistryMole ConceptStoichiometryIdeal Gas Law

Step-by-Step Solution

To find the total volume of hydrogen gas produced, we follow a step-by-step stoichiometric and ideal gas calculation:

Step 1: Write the balanced chemical equation

The reaction between aluminium (Al\text{Al}) and sulfuric acid (H2SO4\text{H}_2\text{SO}_4) is represented by: 2Al (s)+3H2SO4 (aq)→Al2(SO4)3 (aq)+3H2 (g)2\text{Al (s)} + 3\text{H}_2\text{SO}_4\text{ (aq)} \rightarrow \text{Al}_2(\text{SO}_4)_3\text{ (aq)} + 3\text{H}_2\text{ (g)}

Step 2: Calculate the number of moles of each reactant

  1. Moles of Aluminium (Al\text{Al}): nAl=Mass of AlMolar mass of Al=5.4 g27.0 g mol−1=0.2 moln_{\text{Al}} = \frac{\text{Mass of Al}}{\text{Molar mass of Al}} = \frac{5.4\text{ g}}{27.0\text{ g mol}^{-1}} = 0.2\text{ mol}

  2. Moles of Sulfuric Acid (H2SO4\text{H}_2\text{SO}_4): nH2SO4=Molarity×Volume in liters=5.0 M×(50.01000 L)=0.25 moln_{\text{H}_2\text{SO}_4} = \text{Molarity} \times \text{Volume in liters} = 5.0\text{ M} \times \left(\frac{50.0}{1000}\text{ L}\right) = 0.25\text{ mol}

Step 3: Identify the limiting reagent

From the balanced chemical equation, 2 moles of Al2\text{ moles of Al} react with 3 moles of H2SO43\text{ moles of H}_2\text{SO}_4.

Let's check the mole ratio for each reactant:

  • For Al\text{Al}: nAl2=0.22=0.10\frac{n_{\text{Al}}}{2} = \frac{0.2}{2} = 0.10
  • For H2SO4\text{H}_2\text{SO}_4: nH2SO43=0.253≈0.0833\frac{n_{\text{H}_2\text{SO}_4}}{3} = \frac{0.25}{3} \approx 0.0833

Since 0.0833<0.100.0833 < 0.10, H2SO4\text{H}_2\text{SO}_4 is the limiting reagent.

Step 4: Calculate the moles of H2\text{H}_2 gas produced

From the stoichiometric coefficients of the reaction: 3 moles of H2SO4 produce 3 moles of H23\text{ moles of H}_2\text{SO}_4 \text{ produce } 3\text{ moles of H}_2 ⇒nH2=nH2SO4=0.25 mol\Rightarrow n_{\text{H}_2} = n_{\text{H}_2\text{SO}_4} = 0.25\text{ mol}

Step 5: Calculate the volume of H2\text{H}_2 gas generated

Using the ideal gas equation PV=nRTPV = nRT: V=nRTPV = \frac{n R T}{P}

Substitute the given values:

  • n=0.25 moln = 0.25\text{ mol}
  • R=0.082 L atm mol−1 K−1R = 0.082\text{ L atm mol}^{-1}\text{ K}^{-1}
  • T=300 KT = 300\text{ K}
  • P=1.0 atmP = 1.0\text{ atm}

V=0.25 mol×0.082 L atm mol−1 K−1×300 K1.0 atmV = \frac{0.25\text{ mol} \times 0.082\text{ L atm mol}^{-1}\text{ K}^{-1} \times 300\text{ K}}{1.0\text{ atm}} V=0.25×24.6=6.15 LV = 0.25 \times 24.6 = 6.15\text{ L}

Final Answer:

The volume of hydrogen gas generated is 6.15 L6.15\text{ L}.

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