Upgrade to remove ads
Only $2.99/month

Chemistry - Chemical reactions - Decomposition reactions

Terms in this set (11)

NCEA 2011
Students in a laboratory are asked to identify three powders by using thermal decomposition reaction . The powders are copper hydroxide, Cu(OH)2, sodium carbonate, Na2CO3, and sodium hydrogen carbonate, NaHCO3. Explain how you could identify each of these powders by heating them
- Any observations that would be made
- Any tests that would be carried out on products formed to confirm their presence
- Balanced symbol equations for any reactions occuring
Copper hydroxide
Cu(OH)2(s) ---> CuO(s) + H2O(g)
Na2CO3(s) ---> NIL
NaHCO3 ---> Na2CO3(s) + CO2(g) + H2O(g)

- Copper hydroxide is blue and when it is heated the powder seemed to bubble and boil (due to carbon dioxide being given off) and the colour changes to a black powder with steam being produced. To confirm the presence of water, blue cobalt chloride paper can be used and it will go from blue to pink if water is present
- No reaction will occur when sodium carbonate is heated and it will not break down so no colour change will be observed and no gases will be formed
- Sodium hydrogen carbonate is a white powder that will decompose to form a white powder of sodium carbonate. A colourless gas of carbon dioxide is produced as well as steam and the powdered sodium hydrogen carbonate seems to boil and bubble. To test for the presence of carbon dioxide, bubble the gas through limewater. If the limewater goes milky white then carbon dioxide gas is produced. To test for the presence of water, blue cobalt chloride paper can be used, if it goes from blue to pink then this means water is present.
NCEA 2012
In reaction 2, a boiling tube with copper carbonate powder is heated over a Bunsen burner flame.
- Identify the type of reaction that occurs and give a reason for your choice
- Describe any observations that would be made of this reaction, and link these to the substances involved in the reaction
- Outline a test that could be used to confirm the presence of ONE of the products formed
- Balanced symbol equation
- Thermal decomposition reaction
- As a complicated compound of copper carbonate, will be broken down into simpler compounds, of copper oxide and carbon dioxide
- The green carbonate powder would heat and change colour to form a black powder of CuO and a colourless gas of carbon dioxide will be given off. To confirm the presence of carbon dioxide, the gas can be bubbled through limewater and if the limewater turn milky white then carbon dioxide will be present

Balanced symbol equation
CuCO3 ---> CuO + CO2
NCEA 2012
Two test tubes were set up in a school laboratory. Both test tubes were half-filled with identical solutions of hydrogen peroxide (H2O2). It was observed that a few small bubbles of a colourless gas formed on the side of each test tube.

Some solid manganese dioxide (MnO2) was added to test tube B. Describe the observations you would make after manganese dioxide (MnO2) was added to test tube B.
After the manganese dioxide is added to test tube B, a vigorous reaction will take place where a large amount of colourless gas is given off. These will also be bubbling due to the colourless gases being given off. The solution remain colourless. The black solid of manganese dioxide remains
NCEA 2012
Explain the chemistry of this reaction
- Identify the type of reaction that happens in test tube B
- Justify your choice of the type of reaction
- Link the observations made for the reaction in test tube B to the substances involved in the reaction
- Balanced symbol equation for the reaction involving hydrogen peroxide
In test tube B, a catalyst of manganese dioxide is added to it. The catalyst increased the rate of reaction without being used up, by lowering the amount of energy needed for it to occur-thus enables the reaction to happen rapidly in room temperature. Thus in test tube B, the type of reaction is a catalytic decomposition of hydrogen peroxide into water and oxygen. Only a small amount of manganese dioxide will be used as a catalyst doesn't get used up in the reaction. Once the manganese dioxide is put into the colourless solution of hydrogen peroxide, a large amount of oxygen (colourless gas) is released in a short amount of time, the catalyst of manganese dioxide remains after the reaction has occurred and water is also at the bottom of the test tube.

Balanced symbol equation for the reaction involving hydrogen peroxide
2H2O2 ---> 2H2O + O2
NCEA 2013
Two calcium compounds were heated over a bunsen burner in a school laboratory. In the first experiment, calcium carbonate was heated and in the second experiment calcium hydroxide was heated. Compare and contrast the reaction that occurs when solid calcium carbonate is strongly heated, with the reaction that occurs when solid calcium hydroxide is strongly heated.
- Identify the type of reaction that occurs when each substance is strongly heated, and justify your choice
- Describe any observations that would be made in each reaction, and link these observations to the reactants and products involved
- Explain how the gas formed in each reaction could be identified
- balanced symbol equation
- The reaction that occurs when both calcium carbonate and calcium hydroxide are heated is thermal decomposition because heat was applied to both calcium carbonate and calcium hydroxide, both broke down into smaller, simpler species
- When calcium carbonate a white powder is heated over a bunsen burner, another white solid of calcium oxide would form, as well as a colourless gas of carbon dioxide will be given off. To test for the presence of carbon dioxide limewater can be used. Bubble the gas through limewater and if the limewater turns milky white then carbon dioxide is present and has been produced.
- When calcium hydroxide a white solid is heated another white solid of calcium oxide would form as well as a colourless vapour that may condense on the sides of the test tube. This vapour produced is water that can be identified by using blue cobalt chloride paper which will go from blue to pink if water is produced.
- When both calcium carbonate and calcium hydroxide are heated a white solid of calcium oxide is produced because calcium oxide is a smaller, simpler of both calcium carbonate and calcium hydroxide
- But, the gases produced by heating calcium carbonate and calcium hydroxide are different as calcium carbonate releases carbon dioxide which will turn limewater milky white and calcium hydroxide will produce water which will turn blue cobalt chloride paper pink

Balanced symbol equation for heating of calcium carbonate
CaCO3 ---> CaO + CO2

Balanced symbol equation for the heating of calcium hydroxide
Ca(OH)2 ---> CaO + H2O
NCEA 2014
Two decomposition reactions were set up in a laboratory.
Reaction 1 - solid sodium hydrogen carbonate, NaHCO3, was heated over a bunsen burner flame
- Describe any observation that would be made as this reaction occurs and link these to the reactants and products involved in the reaction
- Outline a test that could be used to confirm the presence of one of the products formed
- Balanced symbol equation
- Sodium hydrogen carbonate is a white solid and when heated it breaks down into sodium carbonate a white solid and it bubbles to produce a colourless gas of carbon dioxide and steam if given off
- To confirm the presence of carbon dioxide the gas can be bubbled through limewater, which will turn milky white if carbon dioxide gas if produced

Balanced symbol equation
2NaHCO3 ---> Na2CO3 + CO2
NCEA 2014
Reaction 2 - Test tube 1 and test tube 2 were both half-filled with hydrogen peroxide solution H2O2 some powdered manganese dioxide, MnO2, was then added to test tube 1.
- Compare the observations you would make for the reactions in the two test tubes. Link these observations to the reactants and products involved in the reactions
- Balanced symbol equation
- Compare and contrast the decomposition reactions shown in reaction 1 (sodium hydrogen carbonate, NaHCO3) and reaction 2 (hydrogen peroxide, H2O2)
- Hydrogen peroxide is a colourless liquid, which decomposes slowly to form O2 and H2O, but with manganese dioxide acting as a catalyst it will speed up the reaction and there will be vigorous bubbling to produce O2 gas (a lot of O2 gas is produced in a short amount of time)

Balanced symbol equation
2H2O2 ---> 2H2O + O2

- Both reactions involving breaking down a complicated compound into more simple compounds. As sodium hydrogen carbonate is broken down into sodium carbonate and carbon dioxide and hydrogen peroxide is broken down into water and carbon dioxide. Reaction 1 is an example of thermal decomposition as heat is used for the decomposition of sodium hydrogen carbonate. Whereas reaction 2 is an example of a catalytic decomposition as manganese dioxide is used to speed up the reaction of hydrogen peroxide and is not used up in the reaction
NCEA 2015
Student made samples of copper oxide using three different methods in a school laboratory

Reaction 1 - Copper carbonate (heated)
Reaction 2 - Copper hydroxide (heated)
Reaction 3 - Copper (heated)

- Identify the type of reaction occurring in each experiment
- Describe any observations that would be made during each experiment, and link them to the reactants and products involved
- Explain how the student could identify ONE of the products for each of reactions 1 and 2
Reaction 1 - Thermal decomposition
Reaction 2 - Thermal decomposition
Reaction 3 - Combination

Reaction 1
Copper carbonate is a green solid and is heated to produce Copper oxide a black solid and a colourless gas of carbon dioxide which causes the copper carbonate to jump about.
Reaction 2
Copper hydroxide is a blue solid and when heated it produces copper oxide a black solid and colourless water vapour which causes the initial copper hydroxide to jump about and condenses on the side of the test tube
Reaction 3
Copper is a shiny salmon-pink metal and when heated in oxygen gas it produces a black solid of copper oxide that forms on the surface of the metal

Reaction 1
To test for the presence of carbon dioxide the student could boil the gas through limewater and if it goes milky white then it confirms the presence of carbon dioxide
Reaction 2
To test for the presence of water, blue cobalt chloride paper will change from a blue colour to a pink colour which shows that water has been produced
NCEA 2015
Compare and contrast the three reactions above
- Word and balanced symbol equations
- Explain what is occurring during each of the different reactions
- Where relevant, explain the reaction(s) in terms of electron transfer
Reaction 1 heat
Copper carbonate ---> copper oxide + carbon dioxide
CuCO3 ---> CuO + CO2

Reaction 2 heat
Copper hydroxide ---> copper oxide + water
Cu(OH)2 ---> CuO +H2O

Reaction 3
Copper + oxygen ---> Copper oxide
Cu + 0.5O2 ---> CuO

- In reaction 1 and reaction 2, one substance is broken down into two simpler products. In reaction one copper carbonate when heated breaks down into copper oxide and carbon dioxide. In reaction 2 copper hydroxide when heated breaks down into copper oxide and water. In reaction 3 when the shiny salmon-pink metal is heated in oxygen it produces a compound of copper oxide.

- Only in reaction 3 there is electron transfer. The electrons are transferred from the metal to the non-metal. The magnesium atoms loses 2 electrons to achieve a full stable outer shell and forms the Mg2+ ion. Oxygen gains 2 electrons to achieve a full stable outer shell and forms the negative ion of O2-. The copper ions and oxygen ions are attracted to each other as they are oppositely charged and form an ionic bond and an ionic compound.
NCEA 2016
A small amount of solid manganese dioxide is added to a test tube of freshly prepared hydrogen peroxide solution.
- What observations would be made? Explain your answer by linking any observations to the reactants and products involved.
- What type of reaction is occurring? Explain your answer
- Hydrogen peroxide is a colourless liquid gives off a colourless gas of oxygen which can relit a glowing splint. A colourless solution of water is also produced. The black solid of manganese dioxide stays the same throughout the reaction as it is a catalyst and does not get used up
- This is an example of catalytic decomposition. Hydrogen peroxide breaks down into the more stable and simple products of water and oxygen, with the aid of the catalyst manganese dioxide. The manganese dioxide doesn't take part in the reaction but it speeds up the rate of reaction by providing an alternative pathway with a lower activation energy barrier for the decomposition of hydrogen peroxide to occur
NCEA 2016
Three white solids are knows to be lead hydroxide, sodium hydrogen carbonate and calcium carbonate. How could the three solids be identified using decomposition reactions?
- Balanced symbol equations
Lead hydroxide
- Upon heating lead hydroxide, water and lead oxide will be formed. Condensation will occur on the coldest parts of the test tube. The water vapour turn blue cobalt chloride paper pink confirming the presence of water. The decomposition of lead hydroxide only produces a white solid of lead oxide, and water vapour and therefore can be identified. Limewater will remain colourless when vapour is passed through
- After heating sodium hydrogen carbonate it will produce sodium carbonate a white solid, carbon dioxide a colourless gas and water vapour will be formed. The colourless water vapour will turn blue cobalt chloride paper pink which will confirm the presence of water. The carbon dioxide will turn limewater milky white by producing insoluble calcium carbonate
- Upon heating calcium carbonate, a white solid of calcium oxide and a colourless gas of carbon dioxide will be produced. The carbon dioxide produced will turn limewater milky white and this will confirm the presence of carbon dioxide
- Therefore these solids can be identified by thermal decomposition reactions and cobalt chloride paper and limewater test to confirm the presence of the products produced. The solid which has a decomposition reaction that is positive for cobalt chloride but negative for limewater is lead hydroxide. The solid which upon heating produces positive results for both test if sodium hydrogen carbonate. The solid which produces negative test for cobalt chloride paper but a positive test for limewater is calcium carbonate

Balanced symbol equations
Ph(OH)2 ---> PbO + H2O
2NaHCO3 ---> Na2CO3 + H2O + CO2
CaCO3 ---> CaO + CO2

(Lime water)
Ca(OH)2 + CO2 ---> CaCO3 + H2O
CaCO3 + H2O + CO2 ---> Ca(HCO3)2
THIS SET IS OFTEN IN FOLDERS WITH...