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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

Ammonia is produced from the reaction of nitrogen and hydrogen according to the following balanced equation: $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)$$ a. What is the maximum mass of ammonia that can be produced from a mixture of \(1.00 \times 10^{3} \mathrm{g} \mathrm{N}_{2}\) and \(5.00 \times 10^{2} \mathrm{g} \mathrm{H}_{2} ?\) b. What mass of which starting material would remain unreacted?

Consider the following unbalanced chemical equation for the combustion of pentane \(\left(\mathrm{C}_{5} \mathrm{H}_{12}\right):\) $$\mathrm{C}_{5} \mathrm{H}_{12}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)$$ If \(20.4 \mathrm{g}\) of pentane are burned in excess oxygen, what mass of water can be produced, assuming \(100 \%\) yield?

Adipic acid is an organic compound composed of \(49.31 \%\) C, \(43.79 \% \mathrm{O},\) and the rest hydrogen. If the molar mass of adipic acid is \(146.1 \mathrm{g} / \mathrm{mol},\) what are the empirical and molecular formulas for adipic acid?

The compound \(\mathrm{As}_{2} \mathrm{I}_{4}\) is synthesized by reaction of arsenic metal with arsenic triiodide. If a solid cubic block of arsenic \(\left(d=5.72 \mathrm{g} / \mathrm{cm}^{3}\right)\) that is \(3.00 \mathrm{cm}\) on edge is allowed to react with \(1.01 \times 10^{24}\) molecules of arsenic triiodide, what mass of \(\mathrm{As}_{2} \mathrm{I}_{4}\) can be prepared? If the percent yield of \(\mathrm{As}_{2} \mathrm{I}_{4}\) was \(75.6 \%\) what mass of \(\mathrm{As}_{2} \mathrm{I}_{4}\) was actually isolated?

Hydrogen cyanide is produced industrially from the reaction of gaseous ammonia, oxygen, and methane: $$2 \mathrm{NH}_{3}(g)+3 \mathrm{O}_{2}(g)+2 \mathrm{CH}_{4}(g) \longrightarrow 2 \mathrm{HCN}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$$ If \(5.00 \times 10^{3} \mathrm{kg}\) each of \(\mathrm{NH}_{3}, \mathrm{O}_{2},\) and \(\mathrm{CH}_{4}\) are reacted, what mass of HCN and of \(\mathrm{H}_{2} \mathrm{O}\) will be produced, assuming \(100 \%\) yield?
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