Problem 58 - A Complete Resource Book in Chemistry for JEE Main

Question

The amount of zinc required to produce \(224 \mathrm{~mL}\) of \(\mathrm{H}_{2}\) at STP on treatment with dilute \(\mathrm{H}_{2} \mathrm{SO}_{4}\) will be \((Z n=65)\) (a) \(65.0 \mathrm{~g}\) (b) \(0.65 \mathrm{~g}\) (c) \(6.35 \mathrm{~g}\) (d) \(0.065 \mathrm{~g}\)

Step-by-Step Solution

Answer

The amount of zinc required is 0.65 g.

Step 1: Write the balanced chemical equation

The reaction between zinc (Zn) and dilute sulfuric acid (H\(_2\)SO\(_4\)) produces zinc sulfate (ZnSO\(_4\)) and hydrogen gas (H\(_2\)). The balanced chemical equation is: \[ \text{Zn} + \text{H}_2\text{SO}_4 \rightarrow \text{ZnSO}_4 + \text{H}_2 \] This indicates that 1 mole of Zn produces 1 mole of H\(_2\).

Step 2: Calculate moles of H2 produced

At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters or 22400 mL. Therefore, \(224 \text{ mL}\) of \(\text{H}_2\) corresponds to \(\frac{224}{22400} = 0.01 \text{ moles of } \text{H}_2\).

Step 3: Relate moles of H2 to moles of Zn

Since the reaction equation shows 1 mole of Zn produces 1 mole of H\(_2\), the number of moles of Zn needed to produce 0.01 moles of \(\text{H}_2\) is also 0.01 moles.

Step 4: Calculate mass of Zn required

The molar mass of zinc (Zn) is 65 g/mol. Thus, the mass of zinc required is \( \text{moles of Zn} \times \text{molar mass of Zn} = 0.01 \times 65 = 0.65 \text{ g} \).

Key Concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that involves the calculation of reactants and products in chemical reactions. It can be likened to a recipe in cooking, where exact amounts of ingredients yield specific results. In chemical reactions, stoichiometry helps us determine the exact amount of substances required or produced by using a balanced chemical equation.

In our exercise, stoichiometry is key to finding out how much zinc is needed. With the balanced equation showing that 1 mole of zinc reacts to produce 1 mole of hydrogen gas, stoichiometry helps us relate the moles of one substance to the moles of another. This is done through the use of molar ratios derived from the coefficients in the balanced reaction. Therefore, by knowing the moles of hydrogen gas we wish to produce, we can determine the required moles and thus the mass of zinc needed.

Mole Concept

The mole concept is a central piece to understanding chemical reactions and stoichiometry. A mole is a standard unit of measurement in chemistry that represents a large number of entities. This number is Avogadro's number, approximately \(6.022 \times 10^{23}\), which relates to atoms, molecules, ions, etc.

Utilizing the mole concept in our exercise helps us convert between the mass of a substance and the amount of substance in moles. Because one mole of a gas occupies 22.4 liters at standard temperature and pressure (STP), it becomes straightforward to relate a given volume of gas to moles. In this particular case, 224 mL of hydrogen gas at STP translates to 0.01 moles of hydrogen, as there are 1000 mL in a liter. By applying the mole concept, we can easily bridge the gap between measuring mass and counting particles in chemical equations.

Balanced Chemical Equations

Balanced chemical equations are the backbone for calculating reactants and products in a chemical reaction. A balanced equation accurately represents the conservation of mass and charge by having equal numbers of each type of atom on both sides of the equation. This ensures that matter is neither created nor destroyed during the reaction.

In our reaction involving zinc and sulfuric acid, the balanced chemical equation \(\text{Zn} + \text{H}_2\text{SO}_4 \rightarrow \text{ZnSO}_4 + \text{H}_2\) is crucial. It tells us that one molecule of zinc reacts with one molecule of sulfuric acid to produce one molecule of zinc sulfate and one molecule of hydrogen gas. This 1:1 ratio directly informs the stoichiometric relationships needed to solve for the mass of reactants or products. Without this balance, calculations would yield incorrect and impractical results. Therefore, balancing the equation first is an essential step in any stoichiometric calculation.