smc-4259-50-Balancing Chemical Equations

CHEMISTRY, M2 EQ-Bank 3

Balance the following chemical equations:

$\ce{HCl(aq) + Zn(s) -> ZnCl2(aq) + H2(g)}$ (1 mark)
$\ce{CuSO4(aq) + AgNO3(aq) -> Ag2SO4(s) + Cu(NO3)2(aq)}$ (1 mark)

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a. $\ce{2HCl(aq) + Zn(s) -> ZnCl2(aq) + H2(g)}$

b. $\ce{CuSO4(aq) + 2AgNO3(aq) -> Ag2SO4(s) + Cu(NO3)2(aq)}$

Show Worked Solution

a. $\ce{2HCl(aq) + Zn(s) -> ZnCl2(aq) + H2(g)}$

b. $\ce{CuSO4(aq) + 2AgNO3(aq) -> Ag2SO4(s) + Cu(NO3)2(aq)}$

The following equation represents the reaction of calcium disilicide $\ce{(CaSi2)}$ with antimony trichloride $\ce{(SbCl3)}$ to produce calcium chloride $\ce{(CaCl2)}$, silicon $\ce{(Si)}$, and antimony $\ce{(Sb)}$:

$\ce{a CaSi2 + b SbCl3 -> c Si + d Sb + e CaCl2}$

What are the stoichiometric values for $ a, b, c, d$ and $ e$ in the balanced equation?

$2, 3, 6, 3, 2$
$3, 2, 2, 6, 3$
$3, 3, 6, 2, 3$
$3, 2, 6, 2, 3$

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$D$

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Balancing the chlorine atoms:
$\ce{a CaSi2 + 2SbCl3 -> c Si + d Sb + 3CaCl2}$
Since $\ce{Si}$ and $\ce{Sb}$ are by themselves on the right hand side, they can be easily balanced last.
Next, balance the $\ce{Ca}$ atoms on the left hand side:
$\ce{3CaSi2 + 2SbCl3 -> c Si + d Sb + 3CaCl2}$
The values for $c$ and $d$ can be added to balance the remaining atoms.
$\ce{3CaSi2 + 2SbCl3 -> 6 Si + 2 Sb + 3CaCl2}$

$\Rightarrow D$

CHEMISTRY, M2 EQ-Bank 2 MC

Which is the correct balanced formula equation for the reaction of potassium with water?

$\ce{K(s) + H2O(l) -> KOH(aq) + H2(g)}$
$\ce{2K(s) + 2H2O(aq) -> 2KOH(aq) + H2(g)}$
$\ce{2K(s) + 2H2O(l) -> 2KOH(aq) + H2(g)}$
$\ce{K(s) + 2H2O(aq) -> KOH(aq) + 2H2(g)}$

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$C$

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The correct chemical equation is:
$\ce{2K(s) + 2H2O(l) -> 2KOH(aq) + H2(g)}$
The state of water is always liquid. An aqueous solution refers to a solution where a substance has been dissolved into water.

$\Rightarrow C$

CHEMISTRY, M2 EQ-Bank 1 MC

What numbers are required to correctly balance this equation?

__$\ce{Fe2O3 +}$ __$\ce{CO ->}$ __$\ce{Fe +}$ __$\ce{CO2}$

$1, 3, 2, 3$
$2, 3, 1, 3$
$1, 1, 2, 1$
$2, 4, 2, 4$

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$A$

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Balancing the iron $\ce{Fe}$ atoms:

There are 2 Fe atoms in $\ce{Fe2O3}$, so we place a 2 in front of $\ce{Fe}$ on the product side.
$\ce{Fe2O3 + CO -> 2Fe + CO2}$

Balancing the oxygen $\ce{O}$ atoms:

$\ce{Fe2O3}$ has 3 oxygen atoms. To balance equation, we need 3 $\ce{CO}$ molecules on the reactant side to balance the 3 $\ce{CO2}$ molecules on the product side.
$\ce{Fe2O3 + 3CO -> 2Fe + 3CO2}$

$\Rightarrow A$

CHEMISTRY, M2 EQ-Bank 1

Balance the following chemical equations:

$\ce{C2H6(l) + O2(g) -> H2O(l) + CO2(g)}$ (1 mark)
$\ce{H3PO4(aq) + CuCO3(aq) -> Cu3(PO4)2(aq) + CO2(g) + H2O(l)}$ (1 mark)

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a. $\ce{2C2H6(l) + 7O2(g) -> 6H2O(l) + 4CO2(g)}$

b. $\ce{2H3PO4(aq) + 3CuCO3(aq) -> Cu3(PO4)2(aq) + 3CO2(g) + 3H2O(l)}$

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a. $\ce{2C2H6(l) + 7O2(g) -> 6H2O(l) + 4CO2(g)}$

b. $\ce{2H3PO4(aq) + 3CuCO3(aq) -> Cu3(PO4)2(aq) + 3CO2(g) + 3H2O(l)}$

CHEMISTRY, M2 2013 VCE 8a

In an experiment, 5.85 g of ethanol was ignited with 14.2 g of oxygen.

Write an equation for the complete combustion of ethanol. (1 mark)

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Which reagent is in excess? Calculate the amount, in moles, of the reagent identified as being in excess. (3 marks)

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a. $\ce{C2H5OH(g) + 3O2(g) \rightarrow 2CO2(g) + 3H2O(g) }$

b. $\ce{O2_{\text{(excess)}} = 0.062\ \text{mol}}$

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a. $\ce{C2H5OH(g) + 3O2(g) \rightarrow 2CO2(g) + 3H2O(g) }$

♦ Mean mark 43%.

b. $\ce{MM(C2H5OH) = 2 \times 12.0 + 6 \times 1.0 + 16.0 = 46.0\ \text{g mol}^{-1}}$

$\ce{n(C2H5OH) = \dfrac{5.85}{46.0} = 0.1272\ \text{mol}} $

$\ce{n(O2) = \dfrac{14.2}{32.0} = 0.444\ \text{mol}}$

$\text{Reaction ratio}\ \ \ce{C2H5OH : O2 = 1:3}$

$\Rightarrow \ce{n(O2)_{\text{required}} = 3 \times 0.1272 = 0.382\ \text{mol}\ \ (\ce{O2}\ \text{excess}) }$

$\ce{O2_{\text{(excess)}} = 0.444-0.382 = 0.062\ \text{mol}}$

CHEMISTRY, M2 2014 VCE 11*

Consider the following unbalanced ionic equation.

$\ce{Hg(l) + Cr2O7^2–(aq) + H+(aq)\rightarrow Hg^2+(aq) + Cr^3+(aq) + H2O(l)}$

When this equation is completely balanced including the total charge on each side of the equation, find the coefficient of $\ce{Hg(l)}$. (3 marks)

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$3$

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By balancing the equations by the total number of atoms of each side
$\ce{Hg(l) + Cr2O7^2–(aq) + 14H+(aq)\rightarrow Hg^2+(aq) + 2Cr^3+(aq) + 7H2O(l)}$
The charge on the left side is $+12$ while the charge on the right side is $+8$
The charge on the right hand side needs to be increased by $+4$
This can be done by changing to $\ce{3Hg^2+(aq)}$.
The total balanced equation while following law of conservation of charge is:
$\ce{3Hg(l) + Cr2O7^2–(aq) + 14H+(aq)\rightarrow 3Hg^2+(aq) + 2Cr^3+(aq) + 7H2O(l)}$
Hence the coefficient of $\ce{Hg(l)}$ is $3$.

CHEMISTRY, M2 2015 VCE 3 MC

In an experiment, 0.051 mol of sodium hydroxide, $\ce{NaOH}$, reacted completely with 0.017 mol of citric acid, $\ce{C6H8O7}$.

Which one of the following equations correctly represents the reaction between citric acid and the sodium hydroxide solution?

$\ce{NaOH(aq) + C6H8O7(aq)\rightarrow NaC6H7O7(aq) + H2O(l)}$
$\ce{2NaOH(aq) + C6H8O7(aq)\rightarrow Na2C6H6O7(aq) + 2H2O(l)}$
$\ce{3NaOH(aq) + C6H8O7(aq)\rightarrow Na3C6H5O7(aq) + 3H2O(l)}$
$\ce{4NaOH(aq) + C6H8O7(aq)\rightarrow Na4C6H4O7(aq) + 4H2O(l)}$

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$C$

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$\text{Ratio of moles reacting}\ = 0.051 : 0.017 = 3:1 $

$\ce{n(NaOH) : n(C6H8O7) = 3:1} $

$\ce{3NaOH(aq) + C6H8O7(aq)\rightarrow Na3C6H5O7(aq) + 3H2O(l)}$

$\Rightarrow C$

CHEMISTRY, M2 2022 VCE 3* MC

The correct equation for the incomplete combustion of ethanol is

$\ce{C2H5OH(l) + O2(g)\rightarrow CO(g) + H2O(l)}$
$\ce{C2H5OH(l) + 2O2(g)\rightarrow 2CO(g) + 4H2O(l)}$
$\ce{C2H5OH(l) + 2O2(g)\rightarrow 2CO(g) + 3H2O(l)}$
$\ce{2C2H5OH(l) + 4O2(g)\rightarrow 4CO(g) + 5H2O(l)}$

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$C$

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The products of incomplete combustion are $\ce{CO}$ and $\ce{H2O}$ and correctly balanced as shown in $C$.

$\Rightarrow C$

CHEMISTRY, M2 2020 VCE 6a*

Methane gas, $\ce{CH4}$, can be captured from the breakdown of waste in landfills. $\ce{CH4}$ is also a primary component of natural gas. $\ce{CH4}$ can be used to produce energy through combustion.

Write the balanced equation for the complete combustion of $\ce{CH4}$ to produce carbon dioxide, $\ce{CO2}$ given the unbalanced equation below. (1 mark)

$\ce{CH4(g) + O2 \rightarrow CO2(g) + H2O(l)}$

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$\ce{2CH4(g) + 3O2 \rightarrow 2CO2(g) + 4H2O(l)\ \ \text{or}}$

$\ce{CH4(g) + \dfrac{3}{2}O2 \rightarrow CO2(g) + 2H2O(l)}$

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$\ce{2CH4(g) + 3O2 \rightarrow 2CO2(g) + 4H2O(l)\ \ \text{or}}$

$\ce{CH4(g) + \dfrac{3}{2}O2 \rightarrow CO2(g) + 2H2O(l)}$

CHEMISTRY, M2 2016 VCE 9a

Standard solutions of sodium hydroxide, $\ce{NaOH}$, must be kept in airtight containers. This is because $\ce{NaOH}$ is a strong base and absorbs acidic oxides, such as carbon dioxide, $\ce{CO2}$, from the air and reacts with them. As a result, the concentration of $\ce{NaOH}$ is changed to an unknown extent.

$\ce{CO2}$ in the air reacts with water to form carbonic acid, $\ce{H2CO3}$. This can react with $\ce{NaOH}$ to form sodium carbonate, $\ce{Na2CO3}$.

Write a balanced overall equation for the reaction between $\ce{CO2}$ gas and water to form $\ce{H2CO3}$. (1 mark)

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Write a balanced equation for the complete reaction between $\ce{H2CO3}$ and $\ce{NaOH}$ to form $\ce{Na2CO3}$. (1 mark)

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a. $\ce{CO2(g) + H2O(l) \rightarrow H2CO3(aq)} $

b. $\ce{2NaOH(aq) + H2CO3(aq) \rightarrow Na2CO3(aq) + 2H2O(l)}$

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a. $\ce{CO2(g) + H2O(l) \rightarrow H2CO3(aq)} $

b. $\ce{2NaOH(aq) + H2CO3(aq) \rightarrow Na2CO3(aq) + 2H2O(l)}$

CHEMISTRY, M2 2010 HSC 23a

Write a balanced chemical equation for the complete combustion of 1-butanol. (1 mark)
A student measured the heat of combustion of three different fuels. The results are shown in the table.

\begin{array} {|c|c|}
\hline & \textit{Heat of} \\ \ \ \ \textit{Fuel}\ \ \ & \textit{combustion} \\ & (\text{kJ g}^{-1}) \\
\hline A & -48 \\ B & -38 \\ C & -28 \\
\hline \end{array}

The published value for the heat of combustion of 1-butanol is 2676 kJ mol$^{-1}$.
Which fuel from the table is likely to be 1-butanol? Justify your answer. (2 marks)

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a. $\ce{C4H9OH + 6O2 \rightarrow 4CO2 + 5H2O}$

b. $\text{Convert heat of combustion for each fuel to kJ mol}^{-1}:$

$\ce{MM(C4H9OH) = 12.01 \times 4 + 1.008 \times 9 + 16 + 1.008 = 74.12}$

$\ce{$A$: 48 \times 74 = 3552 kJ mol^{-1}}$

$\ce{$B$: 38 \times 74 = 2812 kJ mol^{-1}}$

$\ce{$C$: 28 \times 74 = 2072 kJ mol^{-1}}$

$\text{Energy will be lost due to heat and fuel impurities.}$
$A\ \text{and}\ B\ \text{have values higher than the published value.}$

$\therefore\ C\ \text{is most likely to be 1-butanol.}$

Show Worked Solution

a. $\ce{C4H9OH + 6O2 \rightarrow 4CO2 + 5H2O}$

b. $\text{Convert heat of combustion for each fuel to kJ mol}^{-1}:$

$\ce{MM(C4H9OH) = 12.01 \times 4 + 1.008 \times 9 + 16 + 1.008 = 74.12}$

$\ce{$A$: 48 \times 74 = 3552 kJ mol^{-1}}$

$\ce{$B$: 38 \times 74 = 2812 kJ mol^{-1}}$

$\ce{$C$: 28 \times 74 = 2072 kJ mol^{-1}}$

$\text{Energy will be lost due to heat and fuel impurities.}$
$A\ \text{and}\ B\ \text{have values higher than the published value.}$

$\therefore\ C\ \text{is most likely to be 1-butanol.}$

CHEMISTRY, M2 2012 HSC 5* MC

Which of the following is a balanced equation representing the fermentation of glucose?

$\ce{C6H12O6}(aq) \rightarrow \ce{C2H5OH}(aq) + \ce{CO2}(g)$
$\ce{C6H12O6}(aq) \rightarrow \ce{2C2H5OH}(aq) + \ce{2CO2}(g)$
$\ce{C6H12O6}(aq) \rightarrow \ce{C2H5OH}(aq) + \ce{2CO2}(g)$
$\ce{2C6H12O6}(aq) \rightarrow \ce{4C2H5OH}(aq) + \ce{4CO2}(g)$

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$B$

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1 glucose molecule undergoes fermentation to produce 2 ethanol molecules and 2 carbon dioxide molecules as seen in $B$.

$=>B$

CHEMISTRY, M2 2013 HSC 22b

A solution contains three cations, $ \ce{Ba}^{2+}, \ce{Cu}^{2+}$ and $\ce{Pb}^{2+}$. The flow chart indicates the plan used to confirm the identity of these cations.

Write a balanced net ionic equation for the formation of Precipitate 1. (2 marks)

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$\ce{Pb^{2+}(aq) + 2Cl-(aq) \rightarrow PbCl2(s)}$

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$\ce{Pb^{2+}(aq) + 2Cl-(aq) \rightarrow PbCl2(s)}$

In the above flow chart, the addition of excess HCl causes one of the cations to precipitate out of the solution.
The remaining two are then distinguished through precipitation with excess sulfuric acid $\ce{(H2SO4)}$.
Of the possible salts, only $\ce{PbCl2(s)}$ is insoluble. Therefore Precipitate 1 is Lead $\text{(II)}$ Chloride.
Balanced net ionic equation: $\ce{Pb^{2+}(aq) + 2Cl-(aq) \rightarrow PbCl2(s)}$

♦♦♦ Mean mark 27%.

CHEMISTRY, M2 2016 HSC 19 MC

Excess barium nitrate solution is added to 200 mL of 0.200 mol L$^{-1}$ sodium sulfate.

What is the mass of the solid formed?

4.65 g
8.69 g
9.33 g
31.5 g

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`C`

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$ \ce{Ba(NO3)2 (aq) + Na2SO4 (aq) \rightarrow BaSO4 (s) + 2NaNO3 (aq)}$

$n(\ce{Na2SO4})=0.2 \times 0.2=0.04\ \text{mol}$

$n(\ce{BaSO4 (s)})=0.04\ \text{mol}$

$m(\ce{BaSO4 (s)})=0.04 \times 233.37=9.33\ \text{g}$

$\Rightarrow C$