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Greeley [361]
3 years ago
7

Write a balanced chemical equation, including physical state symbols, for the decomposition of solid calcium carbonate (CaCO_3)

into solid calcium oxide and gaseous carbon dioxide.
Chemistry
2 answers:
nadya68 [22]3 years ago
6 0

Answer:

CaCO₃(s) → CaO(s) + CO₂(g)

Explanation:

The decomposition reaction always make two compounds from one.

The products always have simpler chemical structure, originated from a determined compound. This can happens  spontaneously or by a third party.

A notable example of decomposition is hydrolysis. As for example the case of water, which decomposes and generates oxygen and hydrogen gas

2H₂O (l) → 2 H₂ (g) + O₂ (g)

In this case, the calium carbonate decomposes into CaO and CO₂

These two, are the products of the decomposition.

Of course, the unique reactant is the Calcium Carbonate

The balanced equation is:

CaCO₃(s) → CaO(s) + CO₂(g)

sweet-ann [11.9K]3 years ago
4 0

Answer:

CaCO3(s) → CaO(s) + CO2(g)

Explanation:

Step 1: Data given

Solid calcium carbonate = CaCO3(s)

solid calcium oxide = CaO(s)

gaseous carbon oxide = CO2(g)

Step 2: The unbalanced equation

CaCO3(s) → CaO(s) + CO2(g)

This equation is already balanced. We don't have to balance anything.

The balanced equation is:

CaCO3(s) → CaO(s) + CO2(g)

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If 12.5 grams of strontium hydroxide is reacted with 150 mL of 3.5 M carbonic acid, identify the limiting reactant.
kiruha [24]

Answer:

Sr(OH)₂ will be the limiting reagent.

Explanation:

First of all, you should know the following balanced chemical equation:

2 H₂CO₃ + 2 Sr(OH)₂ → 4 H₂O + Sr₂(CO₃)₂

The balanced equation is based on the Law of Conservation of Mass, which says that matter cannot be created or destroyed. Therefore, the number of each type of atom on each side of a chemical equation must be the same.      

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, it is possible to use the reaction stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction). By stoichiometry the following amounts in moles react:

  • strontium hydroxide: 2 moles
  • carbonic acid: 2 moles

Now, you know the following masses of the elements:

  • Sr: 87.62 g/mole
  • O: 16 g/mole
  • H: 1 g/mole

So the molar mass of strontium hydroxide is:

Sr(OH)₂= 87.62 g/mole + 2*(16 g/mole + 1 g/mole)= 121.62 g/mole

You apply the following rule of three, if 121.62 grams of hydroxide are present in 1 mole, 12.5 grams in how many moles are they?

moles of strontium hydroxide=\frac{12.5 grams*1 mole}{121.62 grams}

moles of hydroxide= 0.103 moles

On the other hand, you have 150 ml of 3.5 M carbonic acid. Since molarity is the concentration of a solution expressed in the number of moles dissolved per liter of solution, you can apply the following rule of three: if in 1 L there are 3.5 moles of carbonic acid, in 0.150 L (being 1 L = 1000 mL, 0.150 L = 150 mL) how many moles of acid are there?

molesofcarbonicacid=\frac{0.150 L*3.5 moles}{1 L}

moles of carbonic acid= 0.525 moles

Finally, to calculate the limiting reagent, you can use a simple rule of three as follows: if by stoichiometry 2 mole of strontium hydroxide reacts with , how much moles of carbonic acid will be needed if 0.103 moles of strontium hydroxide react?

molesofcarbonicacid=\frac{0.103 moles of strontium hydroxide*2 moles of carbonic acid}{2 moles of strontium hydroxide}

moles of carbonic acid= 0.103 moles

But 0.525 moles are available. Since more moles are available than you need to react with 0.103 moles of strontium hydroxide, <u><em>Sr(OH)₂ will be the limiting reagent.</em></u>

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Explanation:

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