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Chapter 1: Problem 41

What is the total mass of the products formed, when \(51 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{~S}\) is oxidized by oxygen to produce water and sulphur dioxide? (a) \(72 \mathrm{~g}\) (b) \(27 \mathrm{~g}\) (c) \(123 \mathrm{~g}\) (d) \(96 \mathrm{~g}\)

Short Answer

Expert verified
The total mass of the products formed is approximately 123 g.

Step by step solution

01

- Write the balanced chemical equation

First, write down the balanced chemical equation for the reaction where hydrogen sulfide \(\mathrm{H}_{2}S\) is oxidized by oxygen \(O_{2}\) to produce water \(H_{2}O\) and sulfur dioxide \(SO_{2}\). The equation is \(2 \mathrm{H}_{2}S + 3 O_{2} \rightarrow 2 H_{2}O + 2 SO_{2}\).
02

- Calculate moles of \(\mathrm{H}_{2}S\)

Using the molar mass of \(\mathrm{H}_{2}S\) (34.08 g/mol), calculate the number of moles of \(\mathrm{H}_{2}S\) we have from the given mass. This is done by dividing the given mass by the molar mass: \(\frac{51 \mathrm{~g}}{34.08 \mathrm{~g/mol}} = 1.5 \mathrm{~mol} \).
03

- Determine moles of the products

According to the balanced chemical equation, 2 moles of \(\mathrm{H}_{2}S\) produce 2 moles of \(SO_{2}\) and 2 moles of \(H_{2}O\). So, 1.5 moles of \(H_{2}S\) will produce 1.5 moles of \(SO_{2}\) and 1.5 moles of \(H_{2}O\).
04

- Calculate masses of the products

Calculate the total mass of the products using their molar masses. For \(H_{2}O\) (molar mass = 18.02 g/mol), the mass is \(1.5 \mathrm{~mol} \times 18.02 \mathrm{~g/mol} = 27.03 \mathrm{~g}\). For \(SO_{2}\) (molar mass = 64.07 g/mol), the mass is \(1.5 \mathrm{~mol} \times 64.07 \mathrm{~g/mol} = 96.10 \mathrm{~g}\). Adding these gives the total mass: \(27.03 \mathrm{~g} + 96.10 \mathrm{~g} = 123.13 \mathrm{~g} \approx 123 \mathrm{~g}\) (rounded to the nearest whole number).

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Key Concepts

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

Stoichiometry
Stoichiometry is the heart of chemical reactions. It involves the calculation of reactants and products in chemical processes and is an essential tool for chemists and students alike. It helps in figuring out the quantities needed to react completely and the amounts we can expect to produce. For example, consider the baking of a cake - you need a specific recipe to get the desired amount of cake. Stoichiometry is like that recipe for chemical reactions.

In the given exercise, the stoichiometry of the reaction tells us how hydrogen sulfide and oxygen react to form water and sulfur dioxide. The balanced chemical equation provides a molar ratio that guides us through the quantities of each compound involved. For every two moles of hydrogen sulfide, three moles of oxygen react to produce two moles of water and two moles of sulfur dioxide. When working on stoichiometric calculations, always ensure you have a balanced chemical equation first.
Molar Mass Calculation
The molar mass calculation is pivotal in stoichiometry as it connects the mass of a substance to the number of moles. In simple terms, the molar mass is the weight of one mole (6.022 x 1023 particles) of a substance and is expressed in grams per mole (g/mol). This concept allows us to convert between the mass of a substance and the number of moles, facilitating the use of stoichiometry in chemical calculations.

When given a mass of a substance, such as the 51 g of H2S in our exercise, we use the molar mass of hydrogen sulfide, 34.08 g/mol, to find out how many moles we have. By dividing the mass by the molar mass, the exercise demonstrates that 51 g is equivalent to 1.5 moles. Remember, correct determination of molar mass and conversion into moles is crucial for accurately predicting the outcome of chemical reactions.
Oxidation-Reduction Reactions
Oxidation-reduction reactions, or redox reactions, are types of chemical reactions where electrons are transferred between substances. Oxidation refers to the loss of electrons, while reduction means the gain of electrons. These reactions are crucial for many processes, from powering batteries to metabolic pathways in living organisms.

In the context of our exercise, the hydrogen sulfide, H2S, is oxidized by oxygen, O2. The oxygen acts as an oxidizing agent and gains the electrons lost by hydrogen sulfide. Though the detailed electron transfer is not illustrated in the exercise, understanding redox reactions deepens your comprehension of how and why certain substances react together. For a complete understanding, one could consider learning about oxidation numbers and how to determine which substance is oxidized and which is reduced during a reaction.

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Most popular questions from this chapter

Two isotopes of an element \(\mathrm{Q}\) are \(\mathrm{Q}^{97}\) (23.4\% abundance) and \(Q^{94}\) (76.6\% abundance). \(Q^{97}\) is \(8.082\) times heavier than \(\mathrm{C}^{12}\) and \(\mathrm{Q}^{94}\) is \(7.833\) times heavier than \(\mathrm{C}^{12}\). What is the average atomic weight of the element Q? (a) \(94.702\) (b) \(78.913\) (c) \(96.298\) (d) \(94.695\)

When a mixture of aluminium powder and iron (III) oxide is ignited, it produces molten iron and aluminium oxide. In an experiment, \(5.4 \mathrm{~g}\) of aluminium was mixed with \(18.5 \mathrm{~g}\) of iron (III) oxide. At the end of the reaction, the mixture contained \(11.2 \mathrm{~g}\) of iron, \(10.2 \mathrm{~g}\) of aluminium oxide, and an undetermined amount of unreacted iron (III) oxide. No aluminium was left. What is the mass of the iron (III) oxide left? (a) \(2.5 \mathrm{~g}\) (b) \(7.3 \mathrm{~g}\) (c) \(8.3 \mathrm{~g}\) (d) \(2.9 \mathrm{~g}\)

A metal oxide has the formula \(\mathrm{M}, \mathrm{O}_{2}\). It can be reduced by hydrogen to give free metal and water. \(0.1596 \mathrm{~g}\) of the metal oxide required \(6 \mathrm{mg}\) of hydrogen for complete reduction. The atomic mass of the metal is (a) \(111.60\) (b) \(159.60\) (c) \(79.80\) (d) \(55.80\)

When a sample of hydrogen fluoride is cooled to \(303 \mathrm{~K}\), most of the molecules undergo dimerization. If the vapour density of such a sample is 18 , what per cent of total molecules in the sample are in dimer form? \((\mathrm{F}=19)\) (a) \(88.89\) (b) \(80.0\) (c) \(20.0\) (d) \(11.11\)

The fractional abundance of \(\mathrm{Cl}^{35}\) in a sample of chlorine containing only \(\mathrm{Cl}^{35}\) (atomic weight \(=34.9\) ) and \(\mathrm{Cl}^{37}\) (atomic weight \(=36.9\) ) isotopes, is \(0.6\). The average mass number of chlorine is (a) \(35.7\) (b) \(35.8\) (c) \(18.8\) (d) \(35.77\)
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