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Balance the equations: (a) \(\mathrm{Ag}_{2} \mathrm{O} \rightarrow \mathrm{Ag}+\mathrm{O}_{2}\) (b) \(\mathrm{Zn}+\mathrm{HCl}+\mathrm{ZnCl}_{2}+\mathrm{H}_{2}\); (c) \(\mathrm{NaOH}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}+\mathrm{H}_{2} \mathrm{O}\).

Short Answer

Expert verified
The short answer for the balanced equations is as follows: (a) \[\mathrm{Ag}_{2} \mathrm{O} \rightarrow 2 \mathrm{Ag}+\mathrm{O}_{2}\] (b) \[\mathrm{Zn}+2\mathrm{HCl} \rightarrow \mathrm{ZnCl}_{2}+\mathrm{H}_{2}\] (c) \[2\mathrm{NaOH}+\mathrm{H}_{2}\mathrm{SO}_{4} \rightarrow \mathrm{Na}_{2}\mathrm{SO}_{4}+2\mathrm{H}_{2}\mathrm{O}\]

Step by step solution

01

Identify the elements

In this equation, we have two elements: silver (Ag) and oxygen (O).
02

Count the atoms on both sides

On the reactant side, there are 2 silver atoms and 1 oxygen atom. On the product side, there are 1 silver atom and 2 oxygen atoms.
03

Balance the equation

To balance the equation, we need 2 silver atoms and 1 oxygen atom on both sides. Multiplying silver (Ag) by 2 on the product side will give us 2 silver atoms: \[\mathrm{Ag}_{2} \mathrm{O} \rightarrow 2 \mathrm{Ag}+\mathrm{O}_{2}\] The balanced equation is: \[\boxed{\mathrm{Ag}_{2} \mathrm{O} \rightarrow 2 \mathrm{Ag}+\mathrm{O}_{2}}\] (b) Balancing the equation \(\mathrm{Zn}+\mathrm{HCl}+\mathrm{ZnCl}_{2}+\mathrm{H}_{2}\).
04

Identify the elements

In this equation, we have three elements: zinc (Zn), hydrogen (H), and chlorine (Cl).
05

Count the atoms on both sides

On the reactant side, there is 1 zinc atom, 1 hydrogen atom, and 1 chlorine atom. On the product side, there is 1 zinc atom, 2 hydrogen atoms, and 2 chlorine atoms.
06

Balance the equation

We can see that the given equation is not written correctly. To correct it, we must first write the reactants and products correctly, which should be: \[\mathrm{Zn}+\mathrm{HCl} \rightarrow \mathrm{ZnCl}_{2}+\mathrm{H}_{2}\] Now, we need 2 hydrogen atoms and 2 chlorine atoms on both sides. Multiplying HCl by 2 on the reactant side will give us 2 hydrogen atoms and 2 chlorine atoms: \[ \mathrm{Zn}+2\mathrm{HCl} \rightarrow \mathrm{ZnCl}_{_2}+\mathrm{H}_{2}\] The balanced equation is: \[\boxed{\mathrm{Zn}+2\mathrm{HCl} \rightarrow \mathrm{ZnCl}_{2}+\mathrm{H}_{2}}\] (c) Balancing the equation \(\mathrm{NaOH}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{Na}_{2}\mathrm{SO}_{4}+\mathrm{H}_{2} \mathrm{O}\).
07

Identify the elements

In this equation, we have four elements: sodium (Na), hydrogen (H), sulfur (S), and oxygen (O).
08

Count the atoms on both sides

On the reactant side, there is 1 sodium atom, 3 hydrogen atoms, 1 sulfur atom, and 5 oxygen atoms. On the product side, there are 2 sodium atoms, 2 hydrogen atoms, 1 sulfur atom, and 5 oxygen atoms.
09

Balance the equation

We need 2 sodium atoms, 2 hydrogen atoms, 1 sulfur atom, and 4 oxygen atoms on both sides. Creating 2 water molecules on the right side balances the hydrogen and oxygen atoms: \[\mathrm{NaOH}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{Na}_{2}\mathrm{SO}_{4}+2\mathrm{H}_{2} \mathrm{O}\] Now, multiplying NaOH by 2 on the reactant side balances the sodium atoms: \[2\mathrm{NaOH}+\mathrm{H}_{2}\mathrm{SO}_{4} \rightarrow \mathrm{Na}_{2}\mathrm{SO}_{4}+2\mathrm{H}_{2}\mathrm{O}\] The balanced equation is \[\boxed{2\mathrm{NaOH}+\mathrm{H}_{2}\mathrm{SO}_{4} \rightarrow \mathrm{Na}_{2}\mathrm{SO}_{4}+2\mathrm{H}_{2}\mathrm{O}}\]

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Key Concepts

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

Chemical Reaction
A chemical reaction is a process in which one or more substances, known as reactants, are transformed into different substances called products. This transformation is due to the breaking and forming of chemical bonds, and it involves a change in the arrangement of atoms. The law of conservation of mass dictates that in a closed system, mass is neither created nor destroyed in a chemical reaction. Hence, the mass of the reactants equals the mass of the products.

For a better understanding of a chemical reaction, let's consider the given exercise for balancing chemical equations. For example, when silver oxide (´¡²µâ‚‚O) breaks down, it forms silver (Ag) and oxygen gas (°¿â‚‚). The initial unbalanced reaction is written as: \(\text{Ag}_2\text{O} \rightarrow \text{Ag} + \text{O}_2\).The balanced form, adhering to the conservation of mass, appears as: \(\boxed{\text{Ag}_2\text{O} \rightarrow 2\text{Ag} + \text{O}_2}\).

The balancing of equations is crucial because it reflects the actual stoichiometry—the quantitative relationship between reactants and products in a chemical reaction—which must be accurate for the reaction to be correctly represented.
Stoichiometry
Stoichiometry is the section of chemistry that deals with the calculation of the quantities of substances involved in chemical reactions. It is a fundamental concept for chemists because it ensures that equations are balanced in terms of mass and charge, a requirement stemming from the conservation laws. Stoichiometry involves using the coefficients of substances in the balanced chemical equation to determine the relative amounts of reactants and products.

For instance, consider the previously mentioned reaction of sodium hydroxide (NaOH) with sulfuric acid (Hâ‚‚SOâ‚„) in the exercise. The initial equation \(\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{H}_2\text{O}\) is unbalanced. Using stoichiometry, we find that two moles of NaOH react with one mole of Hâ‚‚SOâ‚„ to produce one mole of Naâ‚‚SOâ‚„ and two moles of water (Hâ‚‚O), thus the balanced equation is \(\boxed{2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}}\).

Stoichiometry is not only important in theoretical chemistry but also in practical applications like the formulation of pharmaceutical drugs, where precise amounts of each chemical are crucial.
Chemical Formula
A chemical formula is a concise way of expressing information about the atoms that constitute a particular chemical compound or molecule. It represents the elements present and their relative proportions. Chemical formulas can be simple or empirical, molecular, or structural.
  • An empirical formula shows the simplest whole-number ratio of the atoms in a compound.
  • A molecular formula represents the exact number of atoms of each element in a molecule.
  • A structural formula, which is more complex, also shows the arrangement of atoms in a molecule.

Take the decomposition of silver oxide from the exercise as an example. The chemical formula for silver oxide is ´¡²µâ‚‚O. When it decomposes, it produces silver and oxygen gas, with the chemical formulas Ag and °¿â‚‚, respectively. The correctly balanced equation reveals that two molecules of silver are produced for every one molecule of oxygen gas, demonstrating the stoichiometric relationship that can be understood via the chemical formulas involved.

Understanding chemical formulas is essential in balancing chemical equations since it allows us to easily keep track of the atoms involved in the reactants and products, ensuring that the law of conservation of mass is obeyed in the chemical reaction.

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Most popular questions from this chapter

A metal has an atomic weight of 24 . When it reacts with a non-metal of atomic weight 80, it does so in a ratio of 1 atom to 2 atoms, respectively. With this information, how many grams of non-metal will combine with \(33.3\) g of metal. If \(1 \mathrm{~g}\) of metal is reacted with \(5 \mathrm{~g}\) of non-metal, find the amount of product produced.

How many pounds of air (which is \(23.19 \% \mathrm{O}_{2}\) and \(75.46 \%\) \(\mathrm{N}_{2}\) by weight) would be needed to burn a pound of gasoline by a reaction whereby \(\mathrm{C}_{8} \mathrm{H}_{18}\) reacts with \(\mathrm{O}_{2}\) to form \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) ?

Chromic oxide \(\left(\mathrm{Cr}_{2} \mathrm{O}_{3}\right)\) may be reduced with hydrogen according to the equation \(\mathrm{Cr}_{2} \mathrm{O}_{3}+3 \mathrm{H} 2 \rightarrow 2 \mathrm{Cr}+3 \mathrm{H}_{2} \mathrm{O}\) (a) What weight of hydrogen would be required to reduce g of \(\mathrm{Cr}_{2} \mathrm{O}_{3} ?\) (b) For each mole of metallic chromium prepared, how many moles of hydrogen will be required? (c) What weight of metallic chromium can be prepared from one ton of \(\mathrm{Cr}_{2} \mathrm{O}_{3} ? 1 \mathrm{lb}=454 \mathrm{~g}\).

A chemist reacts ferric sulfate with barium chloride and obtains barium sulfate and ferric chloride. He writes the following balanced equation to express this reaction: \(\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}+3 \mathrm{BaCl}_{2} \rightarrow 3 \mathrm{BaSO}_{4} \downarrow+2 \mathrm{FeCl}_{3}\) (A) How much \(\mathrm{BaC} 1_{2}\) should be used to react with 10 grams of \(\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) ? (B) How much \(\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) will be necessary to produce \(100 \mathrm{~g}\) of \(\mathrm{BaSO}_{4}\) ? (C) From a mixture of \(50 \mathrm{~g}\) of \(\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right) 3\) and \(100 \mathrm{~g}\) of \(\mathrm{BaC} 1_{2}\), how much \(\mathrm{FeC} 1_{3}\) can be produced?

How many liters of phosphine \(\left(\mathrm{PH}_{3}\right)\) gas at STP could be made from \(30 \mathrm{~g}\) of calcium by use of the following sequence of reactions: \(3 \mathrm{Ca}+2 \mathrm{P} \rightarrow \mathrm{Ca}_{3} \mathrm{P}_{2}\) \(\mathrm{Ca}_{3} \mathrm{P}_{2}+6 \mathrm{HCl} \rightarrow 2 \mathrm{PH} 3+3 \mathrm{CaCl}_{2}\) (Molecular weights: \(\mathrm{Ca}=40, \mathrm{PH}_{3}=34\).)

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