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Chapter 3: Problem 104

Balance the following equations: a. $\operatorname{Cr}(s)+\mathrm{S}_{8}(s) \rightarrow \mathrm{Cr}_{2} \mathrm{S}_{3}(s)$ b. \(\operatorname{NaHCO}_{3}(s) \stackrel{\mathrm{Heat}}{\longrightarrow}\) $\mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)$ c. \(\quad \mathrm{KClO}_{3}(s) \stackrel{\mathrm{Heat}}{\longrightarrow}\) \(\mathrm{KCl}(s)+\mathrm{O}_{2}(g)\) d. $\operatorname{Eu}(s)+\mathrm{HF}(g) \rightarrow \operatorname{EuF}_{3}(s)+\mathrm{H}_{2}(g)$

Short Answer

Expert verified
Balanced Equations: a. \(2Cr(s)+3S_8(s) \rightarrow 8Cr_2S_3(s)\) b. \(2NaHCO_3(s) \stackrel{Heat}{\longrightarrow} Na_2CO_3(s)+2CO_2(g)+H_2O(g)\) c. \(2KClO_3(s) \stackrel{Heat}{\longrightarrow} 2KCl(s)+3O_2(g)\) d. \(3Eu(s)+6HF(g) \rightarrow 2EuF_3(s)+3H_2(g)\)

Step by step solution

01

(Identify unbalanced atoms)

Identify the atoms that are unbalanced in the equation: \(Cr(s)+S_8(s) \rightarrow Cr_2S_3(s)\). Chromium (Cr) and sulfur (S) atoms are not balanced.
02

(Balance Chromium atoms)

Begin by balancing chromium (Cr) atoms by placing the appropriate coefficient in front of the reactant species. Currently, there are 2 Cr atoms in the product side, so we need to add a 2 in front of the Cr reactant: \(2Cr(s)+S_8(s) \rightarrow Cr_2S_3(s)\)
03

(Balance Sulfur atoms)

Now, we need to balance sulfur (S) atoms. Place the appropriate coefficient in front of the product species. There are 8 S atoms in the reactant, and 3 in the product. The smallest multiple for both numbers is 24, which means 3 sulfur octamers and 8 chromium sesquisulfide molecules are needed: \(2Cr(s)+3S_8(s) \rightarrow 8Cr_2S_3(s)\) #a.# Balanced Equation: \(2Cr(s)+3S_8(s) \rightarrow 8Cr_2S_3(s)\) #b.
04

(Identify unbalanced atoms)

Identify the atoms that are unbalanced in the equation: \(NaHCO_3(s) \stackrel{Heat}{\longrightarrow} Na_2CO_3(s)+CO_2(g)+H_2O(g)\). Sodium (Na), hydrogen (H), and carbon (C) atoms are not balanced.
05

(Balance Sodium atoms)

Begin by balancing sodium (Na) atoms by placing the appropriate coefficient in front of the reactant species. Currently, there are 2 Na atoms in the product side, so we need to add a 2 in front of the NaHCO_3 reactant: \(2NaHCO_3(s) \stackrel{Heat}{\longrightarrow} Na_2CO_3(s)+CO_2(g)+H_2O(g)\)
06

(Balance Hydrogen atoms)

Now, balance hydrogen (H) atoms by adding the appropriate coefficient in front of the water molecule on the product side. There are 2 H atoms present on both sides, so hydrogen is already balanced.
07

(Balance Carbon atoms)

Lastly, balance the carbon (C) atoms by adding the appropriate coefficient in front of the carbon dioxide molecule on the product side. There are 2 C atoms on the reactant side and only 1 on the product side, therefore a coefficient of 2 is needed: \(2NaHCO_3(s) \stackrel{Heat}{\longrightarrow} Na_2CO_3(s)+2CO_2(g)+H_2O(g)\) #b.# Balanced Equation: \(2NaHCO_3(s) \stackrel{Heat}{\longrightarrow} Na_2CO_3(s)+2CO_2(g)+H_2O(g)\) #c.
08

(Identify unbalanced atoms)

Identify the atoms that are unbalanced in the equation: \(KClO_3(s) \stackrel{Heat}{\longrightarrow} KCl(s)+O_2(g)\). Potassium (K), chlorine (Cl), and oxygen (O) atoms are not balanced.
09

(Balance Potassium and Chlorine atoms)

Since there is one potassium (K) and one chlorine (Cl) atom on both sides of the equation, they are already balanced.
10

(Balance Oxygen atoms)

Balance the oxygen (O) atoms by adding the appropriate coefficient in front of the oxygen molecule on the product side. There are 3 O atoms on the reactant side, and 2 on the product side. The smallest multiple for both numbers is 6, so 2 potassium chlorate molecules and 3 oxygen molecules are needed: \(2KClO_3(s) \stackrel{Heat}{\longrightarrow} 2KCl(s)+3O_2(g)\) #c.# Balanced Equation: \(2KClO_3(s) \stackrel{Heat}{\longrightarrow} 2KCl(s)+3O_2(g)\) #d.
11

(Identify unbalanced atoms)

Identify the atoms that are unbalanced in the equation: \(Eu(s)+HF(g) \rightarrow EuF_3(s)+H_2(g)\). Europium (Eu), hydrogen (H), and fluorine (F) atoms are not balanced.
12

(Balance Europium atoms)

Begin by balancing europium (Eu) atoms by placing the appropriate coefficient in front of the reactant species. Currently, there is 1 Eu atom on both sides, so europium is already balanced.
13

(Balance Hydrogen atoms)

Now, balance hydrogen (H) atoms by adding the appropriate coefficient in front of the hydrogen fluoride reactant molecule. There are 2 H atoms on the product side, so we need to add a 2 in front of the HF reactant: \(Eu(s)+2HF(g) \rightarrow EuF_3(s)+H_2(g)\)
14

(Balance Fluorine atoms)

Lastly, balance fluorine (F) atoms by adding the appropriate coefficient in front of the europium fluoride product molecule. There are 2 F atoms on the reactant side and 3 on the product side. The smallest multiple for both numbers is 6, so 3 europium atoms and 6 hydrogen fluoride molecules are needed: \(3Eu(s)+6HF(g) \rightarrow 2EuF_3(s)+3H_2(g)\) #d.# Balanced Equation: \(3Eu(s)+6HF(g) \rightarrow 2EuF_3(s)+3H_2(g)\)

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Most popular questions from this chapter

A substance \(\mathrm{X}_{2} \mathrm{Z}\) has the composition (by mass) of 40.0\% X and 60.0\(\% \mathrm{Z}\) . What is the composition (by mass) of the compound \(\mathrm{XZ}_{2} ?\)

Hydrogen peroxide is used as a cleansing agent in the treatment of cuts and abrasions for several reasons. It is an oxidizing agent that can directly kill many microorganisms; it decomposes on contact with blood, releasing elemental oxygen gas (which inhibits the growth of anaerobic microorganisms); and it foams on contact with blood, which provides a cleansing action. In the laboratory, small quantities of hydrogen peroxide can be prepared by the action of an acid on an alkaline earth metal peroxide, such as barium peroxide: $$ \mathrm{BaO}_{2}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{BaCl}_{2}(a q) $$ What mass of hydrogen peroxide should result when 1.50 \(\mathrm{g}\) barium peroxide is treated with 88.0 \(\mathrm{mL}\) hydrochloric acid solution containing 0.0272 \(\mathrm{g} \mathrm{HCl}\) per mL? What mass of which reagent is left unreacted?

Para-cresol, a substance used as a disinfectant and in the manufacture of several herbicides, is a molecule that contains the elements carbon, hydrogen, and oxygen. Complete combustion of a \(0.345-\mathrm{g}\) sample of \(p\) -cresol produced 0.983 g carbon dioxide and 0.230 \(\mathrm{g}\) water. Determine the empirical formula for \(p\) -cresol.

Glass is a mixture of several compounds, but a major constituent of most glass is calcium silicate, \(\mathrm{CaSiO}_{3}\) . Glass can be etched by treatment with hydrofluoric acid; HF attacks the calcium silicate of the glass, producing gaseous and water-soluble products (which can be removed by washing the glass). For example, the volumetric glassware in chemistry laboratories is often graduated by using this process. Balance the following equation for the reaction of hydrofluoric acid with calcium silicate. $$ \mathrm{CaSiO}_{3}(s)+\mathrm{HF}(a q) \longrightarrow \mathrm{CaF}_{2}(a q)+\mathrm{SiF}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(l) $$

Maleic acid is an organic compound composed of \(41.39 \% \mathrm{C},\) 3.47$\% \mathrm{H}$ , and the rest oxygen. If 0.129 mole of maleic acid has a mass of 15.0 \(\mathrm{g}\) , what are the empirical and molecular formulas of maleic acid?
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