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(1 point)

Responses

3FeO2
3 cap f e cap o sub 2

Fe2O3
cap f e sub 2 cap o sub 3

2Fe2O3
2 cap f e sub 2 cap o sub 3

4FeO2
4 cap f e cap o sub 2
Question 9
Calcium carbonate (CaCO3
) can undergo a chemical reaction to produce calcium oxide (CaO
)and carbon dioxide. Which possible masses of calcium oxide and carbon dioxide (CO2
) can form when 200g of calcium carbonate undergoes this chemical reaction?(1 point)
Responses

56g CaO
and 44g of CO2
56g cap c A cap o and 44g of cap c cap o sub 2

40g CaO
and 12g of CO2
40g cap c A cap o and 12g of cap c cap o sub 2

200g CaO
and 200g of CO2
200g cap c A cap o and 200g of cap c cap o sub 2

112g CaO
and 88g of CO2
112g cap c A cap o and 88g of cap c cap o sub 2
Question 10
(1 point)
Responses

144g
144g

130g
130g

270g
270g

274g
274g
Question 11
(1 point)
Responses

For every 4 moles of water produced, 3 moles of propane react.
For every 4 moles of water produced, 3 moles of propane react.

For each mole of oxygen that reacts, 3 moles of carbon dioxide are produced.
For each mole of oxygen that reacts, 3 moles of carbon dioxide are produced.

For each mole of propane that reacts, 5 moles of oxygen are produced.
For each mole of propane that reacts, 5 moles of oxygen are produced.

For every 3 moles of carbon dioxide produced, 5 moles of oxygen react.
For every 3 moles of carbon dioxide produced, 5 moles of oxygen react.
Question 12

1 answer

To solve the questions you presented, let's break them down one by one.

Question 9: Reaction of Calcium Carbonate

The reaction of calcium carbonate (CaCO₃) decomposes into calcium oxide (CaO) and carbon dioxide (CO₂) as follows:

\[ \text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2 \]

For every 1 mole of CaCO₃, we produce 1 mole of CaO and 1 mole of CO₂.

Molar Mass Calculations:

  • Molar mass of CaCO₃ = 40 (Ca) + 12 (C) + 3*16 (O) = 100 g/mol
  • Molar mass of CaO = 40 (Ca) + 16 (O) = 56 g/mol
  • Molar mass of CO₂ = 12 (C) + 2*16 (O) = 44 g/mol

From 200 g of CaCO₃:

  • Moles of CaCO₃ = 200 g / 100 g/mol = 2 moles
  • Therefore, moles of CaO produced = 2 moles
  • Therefore, moles of CO₂ produced = 2 moles

Calculating the mass:

  • Mass of CaO = 2 moles * 56 g/mol = 112 g
  • Mass of CO₂ = 2 moles * 44 g/mol = 88 g

So the correct response for the reaction when 200 g of CaCO₃ is decomposed is:

  • 112g CaO and 88g of CO₂

Answer: 112g CaO and 88g of CO₂.

Question 10: (Mass Selection)

It’s unclear what the question regarding mass pertains to, as no particular context or data is provided. However, if this relates to the previous reaction or a general query, without context, it is impossible to directly discern which of the 4 responses (144g, 130g, 270g, 274g) could be the correct answer.

Question 11: Stoichiometry of Propane Combustion

Assuming this is in the context of propane (C₃H₈) combustion, the reaction of propane with oxygen is as follows:

\[ \text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O} \]

From this balanced equation:

  • For every 1 mole of propane consumed, 5 moles of oxygen are used and 3 moles of carbon dioxide are produced.

Thus, regarding the given statements, the true one is:

  • For each mole of propane that reacts, 5 moles of oxygen are consumed.

To summarize:

  1. Question 9 Answer: 112g CaO and 88g of CO₂.
  2. Question 10: Requires context to deduce any specific mass.
  3. Question 11 Answer: For each mole of propane that reacts, 5 moles of oxygen are consumed.