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Chapter 5: Problem 100

Balance each of the following chemical equations. a. \(\mathrm{KO}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{KOH}(a q)+\mathrm{O}_{2}(g)+\mathrm{H}_{2} \mathrm{O}_{2}(a q)\) b. \(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{HNO}_{3}(a q) \rightarrow \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) c. \(\mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{NO}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) d. \(\mathrm{PCl}_{5}(l)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{H}_{3} \mathrm{PO}_{4}(a q)+\mathrm{HCl}(g)\) e. \(\mathrm{CaO}(s)+\mathrm{C}(s) \rightarrow \mathrm{CaC}_{2}(s)+\mathrm{CO}_{2}(g)\) f. \(\operatorname{MoS}_{2}(s)+\mathrm{O}_{2}(g) \rightarrow \operatorname{MoO}_{3}(s)+\mathrm{SO}_{2}(g)\) g. \(\mathrm{FeCO}_{3}(s)+\mathrm{H}_{2} \mathrm{CO}_{3}(a q) \rightarrow \mathrm{Fe}\left(\mathrm{HCO}_{3}\right)_{2}(a q)\)

Short Answer

Expert verified
\(Left: N - 1, O - 2, H - 3\) \(Right: N - 1, O - 3, H - 2\) Step 2: Balance the Hydrogens by placing a coefficient of 2 in front of NH3: \(2 \times NH_3(g) + O_2(g)\) -> \(NO(g) + H_2O(g)\) Step 3: Re-count the number of atoms of each element on both sides of the equation. \(Left: N - 2, O - 2, H - 6\) \(Right: N - 1, O - 3, H - 2\) Step 4: Balance the Nitrogens by placing a coefficient of 2 in front of NO: \(2 \times NH_3(g) + O_2(g)\) -> \(2 \times NO(g) + H_2O(g)\) Step 5: Re-count the number of atoms of each element on both sides of the equation. \(Left: N - 2, O - 2, H - 6\) \(Right: N - 2, O - 4, H - 2\) Step 6: Balance the remaining Oxygens by placing a coefficient of 3 in front of H2O: \(2 \times NH_3(g) + O_2(g)\) -> \(2 \times NO(g) + 3 \times H_2O(g)\) Step 7: Re-count the number of atoms of each element on both sides of the equation. \(Left: N - 2, O - 2, H - 6\) \(Right: N - 2, O - 6, H - 6\) Balanced Equation: \(2 NH_3(g) + O_2(g) \rightarrow 2 NO(g) + 3 H_2O(g)\)

Step by step solution

01

a. Balancing KO2(s) + H2O(l) -> KOH(aq) + O2(g) + H2O2(aq)

: Step 1: Count the number of atoms of each element on both sides of the equation. \(Left: K - 1, O - 3, H - 2\) \(Right: K - 1, O - 5, H - 4\) Step 2: Balance the Oxygens by placing a coefficient of 2 in front of H2O and a coefficient of 1 in front of O2: \(2 \times H_2O(l)\) -> \(KOH(aq) + O_2(g) + H_2O_2(aq)\) Step 3: Re-count the number of atoms of each element on both sides of the equation. \(Left: K - 1, O - 4, H - 4\) \(Right: K - 1, O - 5, H - 4\) Step 4: Balance the remaining Oxygen by placing a coefficient of 2 in front of KO2. \(2 \times KO_2(s) + 2 \times H_2O(l)\) -> \(KOH(aq) + O_2(g) + H_2O_2(aq)\) Step 5: Re-count the number of atoms of each element on both sides of the equation. \(Left: K - 2, O - 6, H - 4\) \(Right: K - 2, O - 6, H - 4\) Balanced Equation: \(2 KO_2(s) + 2 H_2O(l) \rightarrow KOH(aq) + O_2(g) + H_2O_2(aq)\)
02

b. Balancing Fe2O3(s) + HNO3(aq) -> Fe(NO3)3(aq) + H2O(l)

: Step 1: Count the number of atoms of each element on both sides of the equation. \(Left: Fe - 2, O - 6, H - 1, N - 1\) \(Right: Fe - 1, O - 12, H - 2, N - 3\) Step 2: Balance the Irons by placing a coefficient of 2 in front of Fe(NO3)3: \(Fe_2O_3(s) + HNO_3(aq)\) -> \(2 \times Fe(NO_3)_3(aq) + H_2O(l)\) Step 3: Re-count the number of atoms of each element on both sides of the equation. \(Left: Fe - 2, O - 6, H - 1, N - 1\) \(Right: Fe - 2, O - 18, H - 2, N - 6\) Step 4: Balance the Nitrogens by placing a coefficient of 6 in front of HNO3: \(Fe_2O_3(s) + 6 \times HNO_3(aq)\) -> \(2 \times Fe(NO_3)_3(aq) + H_2O(l)\) Step 5: Re-count the number of atoms of each element on both sides of the equation. \(Left: Fe - 2, O - 21, H - 6, N - 6\) \(Right: Fe - 2, O - 18, H - 2, N - 6\) Step 6: Balance the remaining Hydrogens and Oxygens by placing a coefficient of 3 in front of H2O: \(Fe_2O_3(s) + 6 \times HNO_3(aq)\) -> \(2 \times Fe(NO_3)_3(aq) + 3 \times H_2O(l)\) Step 7: Re-count the number of atoms of each element on both sides of the equation. \(Left: Fe - 2, O - 21, H - 6, N - 6\) \(Right: Fe - 2, O - 21, H - 6, N - 6\) Balanced Equation: \(Fe_2O_3(s) + 6 HNO_3(aq) \rightarrow 2 Fe(NO_3)_3(aq) + 3 H_2O(l)\)
03

c. Balancing NH3(g) + O2(g) -> NO(g) + H2O(g)

: Step 1: Count the number of atoms of each element on both sides of the equation.

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Key Concepts

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

Stoichiometry
Understanding stoichiometry is crucial when balancing chemical equations. It involves the calculation of the quantities of reactants and products in chemical reactions. The concept revolves around using coefficients to balance the number of each type of atom on both sides of a chemical equation. These coefficients, which are typically whole numbers, depict the ratio in which chemicals react and form products.
  • Beginners should start by writing down the number of atoms for each element involved in the reaction.
  • Then, adjust the coefficients to get the same number of each type of atom on both sides of the equation.
  • This will ensure that the law of conservation of mass is upheld, maintaining a balance.
By following these steps, students can accurately balance chemical equations and solve stoichiometry problems.
Conservation of Mass
In chemistry, the conservation of mass is a fundamental principle stating that mass is neither created nor destroyed in any chemical reaction. This concept comes into play when balancing chemical equations.
For instance, in any given chemical equation, there must be the same amount of each element on the reactants side and the products side. This ensures that the prediction of how much product gets formed from a given amount of reactant is accurate.
  • If, for example, you start with 2 moles of KO2 in balancing a reaction, you must ensure that all potassium atoms are accounted for in the products.
  • It underscores the permanence and predictability in chemical reactions.
  • Thus, every step involves meticulous counting and adjustments of coefficients to maintain balance.
Following this principle is essential for correctly interpreting chemical reactions and is a cornerstone of stoichiometry.
Chemical Reactions
Understanding chemical reactions is key to mastering any chemistry topic, including equation balancing. Chemical reactions involve the transformation of reactants into products, which are new substances with differing chemical properties.
In equations, reactants are listed on the left, and products on the right, separated by an arrow pointing from left to right.
  • Students should recognize the types of chemical reactions such as synthesis, decomposition, single and double replacement, and combustion.
  • Recognizing the type of reaction can help in predicting the product and balancing the equation rapidly.
  • Every reaction follows specific stoichiometric laws that guide the transformation of atoms from reactants into products.
Once you understand chemical reactions, you will find it easier to balance equations as you will anticipate what to look for and which elements to prioritize during balancing.

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Most popular questions from this chapter

Cumene is a compound containing only carbon and hydrogen that is used in the production of acetone and phenol in the chemical industry. Combustion of \(47.6 \mathrm{mg}\) cumene produces some \(\mathrm{CO}_{2}\) and \(42.8 \mathrm{mg}\) water. The molar mass of cumene is between 115 and \(125 \mathrm{g} / \mathrm{mol} .\) Determine the empirical and molecular formulas.

The molecular formula of acetylsalicylic acid (aspirin), one of the most commonly used pain relievers, is \(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}\) a. Calculate the molar mass of aspirin. b. A typical aspirin tablet contains \(500 . \mathrm{mg} \mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4} .\) What amount (moles) of \(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}\) molecules and what number of molecules of acetylsalicylic acid are in a 500.-mg tablet?

A compound contains only carbon, hydrogen, nitrogen, and oxygen. Combustion of 0.157 g of the compound produced \(0.213 \mathrm{g} \mathrm{CO}_{2}\) and \(0.0310 \mathrm{g} \mathrm{H}_{2} \mathrm{O} .\) In another experiment, it is found that 0.103 g of the compound produces \(0.0230 \mathrm{g} \mathrm{NH}_{3}\) What is the empirical formula of the compound? Hint: Combustion involves reacting with excess \(\mathrm{O}_{2}\). Assume that all the carbon ends up in \(\mathrm{CO}_{2}\) and all the hydrogen ends up in \(\mathrm{H}_{2} \mathrm{O}\). Also assume that all the nitrogen ends up in the \(\mathrm{NH}_{3}\) in the second experiment.

Can the subscripts in a chemical formula be fractions? Explain. Can the coefficients in a balanced chemical equation be fractions? Explain. Changing the subscripts of chemicals can balance the equations mathematically. Why is this unacceptable?

Nitric acid is produced commercially by the Ostwald process, represented by the following equations: $$\begin{array}{c}4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \\\2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) \\\3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g)\end{array}$$ What mass of \(\mathrm{NH}_{3}\) must be used to produce \(1.0 \times 10^{6} \mathrm{kg}\) \(\mathrm{HNO}_{3}\) by the Ostwald process? Assume \(100 \%\) yield in each reaction, and assume that the NO produced in the third step is not recycled.
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