For each of the following reactions, identify the missing reactant(s) or products(s) and then balance the resulting equation. Note that each empty slot may require one or more substances.a. synthesis: ___ ⟢Li2O
b. decomposition:Mg(ClO3)2⟢___, c. double displacement: HNO3+Ca(OH)2β†’___, d. combustion: C5H12+O2⟢___

Respuesta :

Answer:

A. The reactants are= Li and O2

The balanced equation is:

4Li + O2 β€”> 2Li2O

B. The products are: MgCl2 and O2

The balanced equation is:

Mg(ClO3)2 β€”> MgCl2 + 3O2

C. The products are: Ca(NO3)2 and H2O

The balanced equation is:

2HNO3 + Ca(OH)2 β€”> Ca(NO3)2 +

2H2O

D. The products are: CO2 and H2O

The balanced equation is:

C5H12 + 8O2 β€”> 5CO2 + 6H2O

Explanation:

A. ____ β€”> Li2O

The reactants are Li and O2. Thus the equation is given below:

Li + O2 β€”> Li2O

Thus the equation is balanced as follow:

There are 2 atoms of O on the left side and 1 atom on the right side. It can be balance by putting 2 in front of Li2O as shown below:

Li + O2 β€”> 2Li2O

Now, we have 4 atoms of Li on the right side and 1 atom on the left. It can be balance by putting 4 in front of Li as shown below:

4Li + O2 β€”> 2Li2O

Now the equation is balanced

B. Mg(ClO3)2 β€”> __

The products are MgCl2 and O2

The equation is given below:

Mg(ClO3)2 β€”> MgCl2 + O2

The equation can be balance as follow:

There are 6 atoms of O on the left side and 2 atoms on the right side. It can be balance by putting 3 in front of O2 as shown below:

Mg(ClO3)2 β€”> MgCl2 + 3O2

Now the equation is balanced

C. HNO3 + Ca(OH)2 β€”>___

The products are: Ca(NO3)2 and H2O

The equation is given below:

HNO3 + Ca(OH)2 β€”> Ca(NO3)2 + H2O

The equation is balanced as follow:

There are 2 atoms of NO3 on the right side and 1 atom on the left. It can be balance by putting 2 in front of HNO3 as shown below:

2HNO3 + Ca(OH)2 β€”> Ca(NO3)2 + H2O

There are a total of 4 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 2 in front of H2O as shown below:

2HNO3 + Ca(OH)2 β€”> Ca(NO3)2 + 2H2O

Now the equation is balanced

D. C5H12 + O2 β€”>__

The products are: CO2 and H2O

The equation is given below:

C5H12 + O2 β€”> CO2 + H2O

The equation can be balance as follow:

There are 5 atoms of C on the left side and 1 atom on the right side. It can be balance by putting 5 in front of CO2 as shown below:

C5H12 + O2 β€”> 5CO2 + H2O

There are 12 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 6 in front of H2O as shown below:

C5H12 + O2 β€”> 5CO2 + 6H2O

Now, there are a total of 16 atoms of O on the right side and 2 atoms on the left. It can be balance by putting 8 in front of O2 as shown below:

C5H12 + 8O2 β€”> 5CO2 + 6H2O

Now we can see that the equation is balanced.

a. Synthesis: ___ β†’ Liβ‚‚O

Reactants are: Li and Oβ‚‚

Balanced chemical reaction: 4Li + Oβ‚‚ β†’ 2Liβ‚‚O

b. Decomposition :Mg(ClO₃)β‚‚β†’ ___

Products are: MgClβ‚‚ and Oβ‚‚

Balanced Chemical reaction: Mg(ClO₃)β‚‚β†’ MgClβ‚‚ + 3Oβ‚‚

c. Double displacement: HNO₃ + Ca(OH)β‚‚ β†’___

Products are: Ca(NO₃)β‚‚ and Hβ‚‚O

Balanced Chemical reaction: 2HNO₃ + Ca(OH)β‚‚ β†’ Ca(NO₃)β‚‚ + 2Hβ‚‚O

d. Combustion: Cβ‚…H₁₂+Oβ‚‚β†’ ___

Products are: Β COβ‚‚ and Hβ‚‚O

Balanced Chemical reaction: Cβ‚…H₁₂+ 8Oβ‚‚β†’ 5COβ‚‚ + 6Hβ‚‚O

Lets understand these reaction types:

1. Synthesis: Chemical synthesis is the process in which chemical reactions are performed with the idea of converting a reactant into a product or multiple products.

For example: 4Li + Oβ‚‚ β†’ 2Liβ‚‚O

2. Decomposition: In these reactions chemical species break up into simpler parts.

For example: Mg(ClO₃)β‚‚β†’ MgClβ‚‚ + 3Oβ‚‚

3. Double displacement: is a type of chemical reaction where two compounds react, and positive ions and the negative ions of the two reactants switch places, forming two new compounds or products.

For example: Β 2HNO₃ + Ca(OH)β‚‚ β†’ Ca(NO₃)β‚‚ + 2Hβ‚‚O

4. Combustion: These reactions occur when oxygen reacts with another substance and gives off heat and light.

For example: Cβ‚…H₁₂+ 8Oβ‚‚β†’ 5COβ‚‚ + 6Hβ‚‚O

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