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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 \((\mathrm{Zn}=65):\) (a) \(65.0 \mathrm{~g}\) (b) \(0.65 \mathrm{~g}\) (c) \(6.35 \mathrm{~g}\) (d) \(0.065 \mathrm{~g}\)

Short Answer

Expert verified
The amount of zinc required is \(0.65 \mathrm{~g}\).

Step by step solution

01

Understanding the Reaction

The reaction between zinc (\(\text{Zn}\)) and dilute sulfuric acid (\(\text{H}_2\text{SO}_4\)) is given by: \(\text{Zn} + \text{H}_2\text{SO}_4 \rightarrow \text{ZnSO}_4 + \text{H}_2\). This reaction shows that 1 mole of zinc produces 1 mole of hydrogen gas (\(\text{H}_2\)).
02

Volume to Moles at STP

At STP (Standard Temperature and Pressure), 1 mole of any gas occupies \(22,400\, \text{mL}\). So, the number of moles of \(\text{H}_2\) produced is \(\frac{224}{22400} = 0.01\) moles.
03

Determine Moles of Zinc Required

Since 1 mole of \(\text{Zn}\) produces 1 mole of \(\text{H}_2\), the moles of \(\text{Zn}\) required are \(0.01\) moles.
04

Convert Moles of Zinc to Grams

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Reactions
In chemistry, a chemical reaction is a process where reactants are transformed into products. What makes chemical reactions fascinating is how substances interact to change their identities. In the exercise mentioned, zinc (Zn) reacts with sulfuric acid (\(\text{H}_2\text{SO}_4\)) to produce zinc sulfate (ZnSO\(_4\)) and hydrogen gas (\(\text{H}_2\)). This is a classic example of a single displacement reaction.

In this type of reaction:
  • A more reactive metal replaces a less reactive element in a compound.
  • One substance is oxidized, and the other is reduced.
The equation is balanced when each element on the reactant side has an equal number of atoms on the product side. This helps us in stoichiometry to ensure that mass is conserved throughout the reaction.

Visualizing chemical reactions can help improve understanding. For instance, think of the reactants as ingredients in a recipe. Once combined, a reaction occurs, producing something new—a product, just like baking a cake!
Gas Laws at STP
The concept of Standard Temperature and Pressure (STP) is crucial in chemistry, especially when dealing with gases. At STP, the conditions are defined as a temperature of 273 K (0°C) and a pressure of 1 atm.

Why is STP important?
  • It provides a standard reference for scientists to use in calculations involving gases.
  • It makes different chemical data comparable under same conditions.
One remarkable feature of gases at STP is that 1 mole of any gas occupies the same volume: 22,400 mL (22.4 L).

This uniformity is part of what makes gas law calculations straightforward. For example, if we know the volume of gas produced at STP, we can easily determine the amount in moles using the relation: \[ \text{Moles} = \frac{\text{Volume (ml)}}{22,400 \text{ mL/mol}} \]This relationship simplifies the process of converting between volume and moles, making gas reaction calculations more intuitive.
Mole Concept
The mole concept is a fundamental principle in chemistry that allows chemists to relate quantities of substances to their chemical reactions on a macroscopic scale.

Here’s why understanding moles is essential:
  • The mole is like a bridge between the atomic and macroscopic worlds, linking mass, how much of a substance there is, and chemical quantities.
  • 1 mole corresponds to Avogadro's number, approximately \(6.022 \times 10^{23}\) particles (atoms, molecules, etc.).
In the context of our exercise, knowing the number of moles helps us determine the exact amount of zinc needed. By counting moles, we can directly link this quantity to measurable masses using the molar mass of zinc, which is given as 65 g/mol.

This helps in converting from moles to grams: \[ \text{Mass (g)} = \text{Moles} \times \text{Molar Mass (g/mol)} \]Ultimately, the mole concept is key to quantify and convert between substances accurately, underpinning all stoichiometric calculations in chemistry.

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Most popular questions from this chapter

Sodium bicarbonate on heating decomposes to form sodium carbonate, \(\mathrm{CO}_{2}\) and water. If \(0.2\) moles of sodium bicarbonate is completely decomposed, how many moles of sodium carbonate is formed? (a) \(0.1\) (b) \(0.2\) (c) \(0.05\) (d) \(0.025\)

What is the volume (in litres) of oxygen at STP required for complete combustion of \(32 \mathrm{~g}\) of \(\mathrm{CH}_{4}\) ? (mol. wt. of \(\mathrm{CH}_{4}=16\) ) (a) \(89.6\) (b) \(189.6\) (c) \(98.4\) (d) \(169.5\)

The weight of one molecule of a compound \(\mathrm{C}_{60} \mathrm{H}_{122}\) is: (a) \(1.3 \times 10^{-20} \mathrm{~g}\) (b) \(5.01 \times 10^{-21} \mathrm{~g}\) (c) \(3.72 \times 10^{23} \mathrm{~g}\) (d) \(1.4 \times 10^{-21} \mathrm{~g}\)

The reaction, \(2 \mathrm{Al}(\mathrm{s})+6 \mathrm{HCl}\) (aq) \(\longrightarrow 2 \mathrm{Al}^{3+}(\mathrm{aq})\) \(+6 \mathrm{Cl}^{-}(\mathrm{aq})+3 \mathrm{H}_{2}(\mathrm{~g})\) (a) \(33.6 \mathrm{~L} \mathrm{H}_{2}\) (g) is produced regardless of temperature and pressure for every mole of Al that reacts (b) \(67.2 \mathrm{~L} \mathrm{H}_{2}(\mathrm{~g})\) at STP is produced for every mole of Al that reacts (c) \(11.2 \mathrm{~L} \mathrm{H}_{2}(\mathrm{~g})\) at STP is produced for every mole of HCl (aq) consumed (d) \(6 \mathrm{~L}\) HCl (aq) is consumed for every \(3 \mathrm{~L} \mathrm{H}_{2}(\mathrm{~g})\) produced

The incorrect statement for \(14 \mathrm{~g}\) of \(\mathrm{CO}\) is: (a) It occupies \(2.24\) litre at NTP (b) It corresponds to \(0.5 \mathrm{~mol}\) of \(\mathrm{CO}\) (c) It corresponds to same mol of \(\mathrm{CO}\) and \(\mathrm{N}_{2}\) (d) It corresponds to \(3.01 \times 10^{23}\) molecules of \(\mathrm{CO}\)

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