Writing and Balancing Chemical Equations Quiz
The question sheet
Reveal any answer as you study-
A balanced chemical equation is required to be consistent with which law?
- Law of definite proportions
- Law of constant composition
- Law of multiple proportions
- Law of conservation of matter
Reveal answer
Answer: Law of conservation of matter
Source evidence
PDF page 353: The chemical equation described in section 4.1 is balanced, meaning that equal numbers of atoms for each element involved in the reaction are represented on the reactant and product sides. This is a requirement the equation must satisfy to be consistent with the law of conservation of matter. It may be confirmed by simply summing the numbers of atoms on either side of the arrow and comparing these sums to ensure they are equal. Note that the number of atoms for a given element is calculated by multiplying the coefficient of any formula containing that element by the element’s subscript in the formula. If an element appears in more than one formula on a given side of the equation, the number of atoms represented in each must be computed and then added together. For example, both product species in the example reaction, CO2 and H2O, contain the element oxygen, and so the number of oxygen atoms on the product side of the equation is ⎛ 2 O atoms ⎞ ⎛ 1 O atom ⎞
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How is the number of atoms of a given element in a formula calculated?
- Adding coefficient and subscript
- Subscript minus coefficient
- Dividing subscript by coefficient
- Coefficient times subscript
Reveal answer
Answer: Coefficient times subscript
Source evidence
PDF page 353: The chemical equation described in section 4.1 is balanced, meaning that equal numbers of atoms for each element involved in the reaction are represented on the reactant and product sides. This is a requirement the equation must satisfy to be consistent with the law of conservation of matter. It may be confirmed by simply summing the numbers of atoms on either side of the arrow and comparing these sums to ensure they are equal. Note that the number of atoms for a given element is calculated by multiplying the coefficient of any formula containing that element by the element’s subscript in the formula. If an element appears in more than one formula on a given side of the equation, the number of atoms represented in each must be computed and then added together. For example, both product species in the example reaction, CO2 and H2O, contain the element oxygen, and so the number of oxygen atoms on the product side of the equation is ⎛ 2 O atoms ⎞ ⎛ 1 O atom ⎞
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When balancing equations, what may NOT be changed without altering a substance's identity?
- Coefficients
- Physical state labels
- Formula subscripts
- The reaction arrow
Reveal answer
Answer: Formula subscripts
Source evidence
PDF page 354: The numbers of H atoms on the reactant and product sides of the equation are equal, but the numbers of O atoms are not. To achieve balance, the coefficients of the equation may be changed as needed. Keep in mind, of course, that the formula subscripts define, in part, the identity of the substance, and so these cannot be changed without altering the qualitative meaning of the equation. For example, changing the reactant formula from H2O to H2O2 would yield balance in the number of atoms, but doing so also changes the reactant’s identity (it’s now hydrogen peroxide and not water). The O atom balance may be achieved by changing the coefficient for H2O to 2. 2H O ⟶ H + O (unbalanced) 2 2 2
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What is the balanced equation for the decomposition of water?
- 2H2O → H2 + 2O2
- 2H2O → 2H2 + O2
- H2O → H2 + O2
- H2O2 → H2 + O2
Reveal answer
Answer: 2H2O → 2H2 + O2
Source evidence
PDF page 354: These coefficients yield equal numbers of both H and O atoms on the reactant and product sides, and the balanced equation is, therefore: 2H O ⟶ 2H + O 2 2 2
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What is the balanced equation for N2 and O2 forming dinitrogen pentoxide?
- 2N2 + 5O2 → 2N2O5
- N2 + O2 → N2O5
- N2 + 5O2 → 2N2O5
- 2N2 + 5O2 → N2O5
Reveal answer
Answer: 2N2 + 5O2 → 2N2O5
Source evidence
PDF page 355: The N atom balance has been upset by this change; it is restored by changing the coefficient for the reactant N2 to 2. 2N + 5O ⟶ 2N O 2 2 2 5 Element Reactants Products Balanced? N 4 = 4, yes
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What is the balanced equation for ethane (C2H6) reacting with oxygen?
- 2C2H6 + 7O2 → 3H2O + 4CO2
- C2H6 + O2 → H2O + CO2
- C2H6 + 7O2 → 3H2O + 2CO2
- 2C2H6 + 7O2 → 6H2O + 4CO2
Reveal answer
Answer: 2C2H6 + 7O2 → 6H2O + 4CO2
Source evidence
PDF page 356: 7 with the O2 reactant to yield an odd number, so a fractional coefficient, , is used instead to yield a provisional 2 balanced equation: 7 C H + O ⟶ 3H O + 2CO 2 2 2 2 6 2 A conventional balanced equation with integer-only coefficients is derived by multiplying each coefficient by 2: 2C H + 7O ⟶ 6H O + 4CO 2 2 2 2 6 Finally with regard to balanced equations, recall that convention dictates use of the smallest whole-number coefficients. Although the equation for the reaction between molecular nitrogen and molecular hydrogen to produce ammonia is, indeed, balanced, 3N + 9H ⟶ 6NH 2 2 3 the coefficients are not the smallest possible integers representing the relative numbers of reactant and product molecules. Dividing each coefficient by the greatest common factor, 3, gives the preferred equation: N + 3H ⟶ 2NH 2 2 3
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Why can fractional coefficients be used while balancing an equation?
- They stay in the final equation
- They can be multiplied to integers later
- They change substance identity
- They cancel out oxygen
Reveal answer
Answer: They can be multiplied to integers later
Source evidence
PDF page 355: It is sometimes convenient to use fractions instead of integers as intermediate coefficients in the process of balancing a chemical equation. When balance is achieved, all the equation’s coefficients may then be multiplied by a whole number to convert the fractional coefficients to integers without upsetting the atom balance. For example, consider the reaction of ethane (C2H6) with oxygen to yield H2O and CO2, represented by the unbalanced equation: C H + O ⟶ H O + CO (unbalanced) 2 2 2 2 6 Following the usual inspection approach, one might first balance C and H atoms by changing the coefficients for the two product species, as shown: C H + O ⟶ 3H O + 2CO (unbalanced) 2 2 2 2 6 This results in seven O atoms on the product side of the equation, an odd number—no integer coefficient can be used
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By convention, chemical equations should use which coefficients?
- Only even numbers
- Largest possible integers
- Only fractional coefficients
- Smallest whole-number coefficients
Reveal answer
Answer: Smallest whole-number coefficients
Source evidence
PDF page 356: 7 with the O2 reactant to yield an odd number, so a fractional coefficient, , is used instead to yield a provisional 2 balanced equation: 7 C H + O ⟶ 3H O + 2CO 2 2 2 2 6 2 A conventional balanced equation with integer-only coefficients is derived by multiplying each coefficient by 2: 2C H + 7O ⟶ 6H O + 4CO 2 2 2 2 6 Finally with regard to balanced equations, recall that convention dictates use of the smallest whole-number coefficients. Although the equation for the reaction between molecular nitrogen and molecular hydrogen to produce ammonia is, indeed, balanced, 3N + 9H ⟶ 6NH 2 2 3 the coefficients are not the smallest possible integers representing the relative numbers of reactant and product molecules. Dividing each coefficient by the greatest common factor, 3, gives the preferred equation: N + 3H ⟶ 2NH 2 2 3
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Simplifying 3N2 + 9H2 → 6NH3 to smallest integers gives what?
- N2 + H2 → NH3
- N2 + 3H2 → 2NH3
- 3N2 + 9H2 → 6NH3
- 2N2 + 3H2 → 2NH3
Reveal answer
Answer: N2 + 3H2 → 2NH3
Source evidence
PDF page 356: 7 with the O2 reactant to yield an odd number, so a fractional coefficient, , is used instead to yield a provisional 2 balanced equation: 7 C H + O ⟶ 3H O + 2CO 2 2 2 2 6 2 A conventional balanced equation with integer-only coefficients is derived by multiplying each coefficient by 2: 2C H + 7O ⟶ 6H O + 4CO 2 2 2 2 6 Finally with regard to balanced equations, recall that convention dictates use of the smallest whole-number coefficients. Although the equation for the reaction between molecular nitrogen and molecular hydrogen to produce ammonia is, indeed, balanced, 3N + 9H ⟶ 6NH 2 2 3 the coefficients are not the smallest possible integers representing the relative numbers of reactant and product molecules. Dividing each coefficient by the greatest common factor, 3, gives the preferred equation: N + 3H ⟶ 2NH 2 2 3
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Which abbreviation indicates a substance dissolved in water?
- l
- s
- g
- aq
Reveal answer
Answer: aq
Source evidence
PDF page 356: The physical states of reactants and products in chemical equations very often are indicated with a parenthetical abbreviation following the formulas. Common abbreviations include s for solids, l for liquids, g for gases, and aq for substances dissolved in water (aqueous solutions, as introduced in the preceding chapter). These notations are illustrated in the example equation here: 2Na(s) + 2H O(l) ⟶ 2NaOH(aq) + H (g) 2 2 This equation represents the reaction that takes place when sodium metal is placed in water. The solid sodium reacts with liquid water to produce molecular hydrogen gas and the ionic compound sodium hydroxide (a solid in pure form, but readily dissolved in water). Special conditions necessary for a reaction are sometimes designated by writing a word or symbol above or below the equation’s arrow. For example, a reaction carried out by heating may be indicated by the uppercase Greek letter delta (Δ) over the arrow. Δ CaCO (s) ⟶ CaO(s) + CO (g) 3 2 Other examples of these special conditions will be encountered in more depth in later chapters.
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What does the Greek letter delta (Δ) over a reaction arrow indicate?
- A reaction carried out by heating
- A gas is produced
- A reversible reaction
- An aqueous solution
Reveal answer
Answer: A reaction carried out by heating
Source evidence
PDF page 356: The physical states of reactants and products in chemical equations very often are indicated with a parenthetical abbreviation following the formulas. Common abbreviations include s for solids, l for liquids, g for gases, and aq for substances dissolved in water (aqueous solutions, as introduced in the preceding chapter). These notations are illustrated in the example equation here: 2Na(s) + 2H O(l) ⟶ 2NaOH(aq) + H (g) 2 2 This equation represents the reaction that takes place when sodium metal is placed in water. The solid sodium reacts with liquid water to produce molecular hydrogen gas and the ionic compound sodium hydroxide (a solid in pure form, but readily dissolved in water). Special conditions necessary for a reaction are sometimes designated by writing a word or symbol above or below the equation’s arrow. For example, a reaction carried out by heating may be indicated by the uppercase Greek letter delta (Δ) over the arrow. Δ CaCO (s) ⟶ CaO(s) + CO (g) 3 2 Other examples of these special conditions will be encountered in more depth in later chapters.
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An equation that does not explicitly represent ionic species in solution is called a:
- Net ionic equation
- Complete ionic equation
- Spectator equation
- Molecular equation
Reveal answer
Answer: Molecular equation
Source evidence
PDF page 357: CaCl (aq) + 2AgNO (aq) ⟶ Ca(NO ) (aq) + 2AgCl(s) 2 3 3 2 This balanced equation, derived in the usual fashion, is called a molecular equation because it doesn’t explicitly represent the ionic species that are present in solution. When ionic compounds dissolve in water, they may dissociate into their constituent ions, which are subsequently dispersed homogenously throughout the resulting solution (a thorough discussion of this important process is provided in the chapter on solutions). Ionic compounds dissolved in water are, therefore, more realistically represented as dissociated ions, in this case:
Chemistry: Atoms First
Chemistry: Atoms First by OpenStax, used under CC BY 4.0. Changes made by Stratacademy.
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