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CHEMISTRY, M2 EQ-Bank 2

In an experiment, calcium carbonate \(\ce{(CaCO3)}\) is heated strongly to produce calcium oxide \(\ce{(CaO)}\) and carbon dioxide according to the reaction below:

\(\ce{CaCO3(s) -> CaO(s) + CO2(g)}\)

A student starts with 50.0 g of calcium carbonate. After heating, they collect 28.0 g of calcium oxide.

  1. Using the law of conservation of mass, calculate the mass of carbon dioxide gas produced in this reaction.   (2 marks)
  1. Explain how the law of conservation of mass applies to this reaction.   (2 marks)
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a.    \(22.0\ \text{g}\)

b.    The law of conservation of mass states that the mass of reactants in a chemical reaction must equal the mass of the products.

→ Here, the 50.0 g of calcium carbonate decomposes into 28.0 g of calcium oxide and 22.0 g of carbon dioxide gas.

→ The total mass of products (28.0 g + 22.0 g) equals the initial mass of reactants (50.0 g), confirming that mass is conserved in this reaction.

Show Worked Solution

a.    Law of conservation of mass:

→ The total mass of reactants must equal the total mass of products.

\(m\ce{(CO2)}\) \(=m\ce{(CaCO3)}-m\ce{(CaO)}\)  
  \(= 50.0-28.0\)  
  \(=22.0\ \text{g}\)  

 

b.    Application of law to reaction:

→ The mass of reactants in a chemical reaction must equal the mass of the products.

→ Here, the 50.0 g of calcium carbonate decomposes into 28.0 g of calcium oxide and 22.0 g of carbon dioxide gas.

→ The total mass of products (28.0 g + 22.0 g) equals the initial mass of reactants (50.0 g), confirming that mass is conserved in this reaction.

CHEMISTRY, M2 2012 VCE 14*

A desalination plant produces 200 gigalitres (GL) of fresh water each year. The maximum level of boron permitted in desalinated water is 0.5 ppm (0.5 mg L\(^{-1}\)).

Calculate the maximum mass, in kilograms, of boron that is permitted in one year's production of desalinated water.   (2 marks)

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\(1.0 \times 10^{5}\ \text{kg}\)

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\(\text{Volume of water} = 200 \times 10^{9}\ \text{L} \)

\(\ce{m(B)}\) \(=0.5\ \text{mg L}^{-1} \times 200 \times 10^{9}\ \text{L} \)  
  \(=100 \times 10^{9}\ \text{mg} \)  
  \(=100 \times 10^{6}\ \text{g} \)  
  \(=100 \times 10^{3}\ \text{kg} \)  
  \(=1.0 \times 10^{5}\ \text{kg} \)  
♦ Mean mark 49%.
COMMENT: Multiplying \(\ce{m(B)}\) by Boron’s molar mass was a common error.

CHEMISTRY, M2 2006 HSC 10 MC

Phosphorus pentoxide reacts with water to form phosphoric acid according to the following equation.

\(\ce{P2O5}(s) + \ce{3H2O}(l) \rightarrow \ce{2H3PO4}(aq)\)

Phosphoric acid reacts with sodium hydroxide according to the following equation.

\(\ce{H3PO4}(aq) + \ce{3NaOH}(aq) \rightarrow \ce{Na3PO4}(aq) + \ce{3H2O}(l)\)

A student reacted 1.42 g of phosphorus pentoxide with excess water.

What volume of 0.30 mol L\(^{-1}\) sodium hydroxide would be required to neutralise all the phosphoric acid produced?

  1. 0.067 L
  2. 0.10 L
  3. 0.20 L
  4. 5.0 L
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\(C\)

Show Worked Solution
\(MM(\ce{P2O5})\) \(=2(30.97)+5(16)\)  
  \(=141.94\ \text{g\mol}\)  

 

\(n(\ce{P2O5})=\dfrac{1.42}{141.94}=0.01 \ \text{mol}\)

 \(n(\ce{H3PO4}) = 2 \times 0.01 = 0.02\ \text{mol}\)

 \(n(\ce{NaOH})\) for neutralisation \(= 3 \times 0.02 = 0.06\ \text{mol}\)

\(c\) \(=\dfrac{n}{V}\)  
\(V\) \(=\dfrac{n}{c}\)  
  \(=\dfrac{0.06}{0.3}\)  
  \(=0.2\ \text{L}\)  

 

\( \Rightarrow C\)

CHEMISTRY, M2 2011 HSC 20 MC

When charcoal reacts in the presence of oxygen, carbon monoxide and carbon dioxide are produced according to the following chemical reactions.

\( \ce{C}(s) +\frac{1}{2} \ce{O}_2(g) \rightarrow \mathrm{CO}(g)\)

\(\ce{C}(s) +\ce{O2}(g) \rightarrow \ce{CO2}(g)\)

What would be the total mass of gas produced when 400 g of charcoal is reacted, assuming equal amounts are consumed in each reaction?

  1. 0.93 kg
  2. 1.2 kg
  3. 1.5 kg
  4. 2.5 kg
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\(B\)

Show Worked Solution

\(n(\ce{C} (s))=\dfrac{400}{12.01}=33.3\ \text{mol}\)

→ 16.65 \(\text{mol}\) of charcoal goes into each reaction. Therefore 16.65 \(\text{mol}\) of \(\ce{CO}(g)\) and \(\ce{CO2}(g)\) are produced.

\(m(\ce{CO})=16.65 \times 28.01 =466.4\ \text{g}\)

\(m(\ce{CO2})=16.65 \times 44.01 =732.8\ \text{g}\)

→ Total mass produced = 1199.2 \(\text{g}\) = 1.2 \(\text{kg}\).

\(\Rightarrow B\)

CHEMISTRY, M2 2014 HSC 25b*

Under conditions of low oxygen levels, octane can undergo incomplete combustion according to the following chemical equation:

\( \ce{2C8H18}(l) + \ce{17O2}(g) \rightarrow \ce{6C}(s)+4 \ce{CO}(g) + \ce{6CO2}(g) + \ce{18H2O}(l)\)

Calculate the mass of soot \((\ce{C}(s))\) produced if 50 grams of octane are combusted in this way with 30 grams of oxygen and the mass of soot accounts for \(\dfrac{1}{5}\) of the total mass of the products.  (2 marks)

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\(16\ \text{g}\)

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→ Total mass of the reactants 80 grams.

→ By the law of conservation of mass, the total mass of the products will be 80 grams.

→ Mass of soot \(=\dfrac{1}{5} \times 80 = 16\ \text{g}\)

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?

  1. 4.65 g
  2. 8.69 g
  3. 9.33 g
  4. 31.5 g
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`C`

Show Worked Solution

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

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

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

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

\(\Rightarrow C\)