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IgorC [24]
3 years ago
11

List the number of each type of atom on the left side of the equation C5H12(g)+8O2(g)→5CO2(g)+6H2O(g) C 5 H 12 ( g ) + 8 O 2 ( g

) → 5 C O 2 ( g ) + 6 H 2 O ( g ) Enter your answers separated by commas (the order of the numbers is the same as the order of the elements on the left side of the equation).
Chemistry
2 answers:
Gala2k [10]3 years ago
5 0

Answer:

Carbon=5, hydrogen=12, oxygen=16

Explanation:

Carbon=5, hydrogen=12, oxygen=16

In order to effectively count the number of atoms, we look at the equation closely and take note of the stoichiometric coefficients of each reactant as this influences the number of atoms of that element present.

For instance, oxygen is diatomic and has a stoichiometric coefficient of 8. This implies the there are sixteen atoms of oxygen altogether.

Note that the left hand side refers to the reactants side.

slega [8]3 years ago
4 0

Answer:

Explanation:

Equation of the reaction:

C5H12(g) + 8O2(g) → 5CO2(g) + 6H2O(g)

1. Ensure the equation of the reaction is balanced properly.

2. Pick any side of the reaction a d List the atoms present.

Carbon, C = 5 atoms

Oxygen, O = (8 × 2)

= 16 atoms

Hydrogen, H = 5 atoms

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Please help me equalize: PbO2 + MnSO4 + HNO3 = HMnO4 + PbSO4 + Pb(NO3)2 + H2O
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Answer:

5PbO₂ + 2MnSO₄ + 6HNO₃ ⟶ 2PbSO₄ + 3Pb(NO₃)₂ + 2HMnO₄+ 2H₂O

Explanation:

PbO₂ + MnSO₄ + HNO₃ ⟶ HMnO₄ + PbSO₄ + Pb(NO₃)₂ + H₂O

It will be easiest to balance this equation by the ion-electron method.

1. Write the ionic equation

PbO₂ + Mn²⁺ + SO₄²⁻ + H⁺ + NO₃⁻ ⟶ H⁺ + MnO₄⁻ + Pb²⁺ + SO₄²⁻ + Pb²⁺ + NO₃⁻ + H₂O

2. Eliminate H⁺, H₂O, and spectator ions

PbO₂ + Mn²⁺ ⟶ MnO₄⁻ + Pb²⁺  

3. Separate the skeleton equation into two half-reactions.

PbO₂  ⟶ Pb²⁺  

Mn²⁺ ⟶ MnO₄⁻

4. Balance all atoms other than H and O

Done

5. Balance O by adding water molecules to the deficient side

           PbO₂  ⟶ Pb²⁺ + 2H₂O

Mn²⁺ + 4H₂O ⟶ MnO₄⁻

6. Balance H by adding H⁺ ions to the deficient side.

  PbO₂+ 4H⁺ ⟶ Pb²⁺ + 2H₂O

Mn²⁺ + 4H₂O ⟶ MnO₄⁻ + 8H⁺

7. Balance charge by adding electrons to the deficient side.

PbO₂+ 4H⁺ + 2e⁻ ⟶ Pb²⁺ + 2H₂O

     Mn²⁺ + 4H₂O ⟶ MnO₄⁻ + 8H⁺ + 5e-

 

8. Multiply each half-reaction by a number to equalize the electrons transferred.

5 × [PbO₂+ 4H⁺ + 2e⁻ ⟶ Pb²⁺ + 2H₂O]

     2 × [Mn²⁺ + 4H₂O ⟶ MnO₄⁻ + 8H⁺ + 5e⁻]

9. Add the two half-reactions.

                          5PbO₂+ 20H⁺ + 10e⁻ ⟶ 5Pb²⁺ + 10H₂O

<u>                                    2Mn²⁺ + 8H₂O ⟶ 2MnO₄⁻ + 16H⁺ + 10e⁻                    </u>

5PbO₂ + 2Mn² + 8H₂O + 20H⁺ + 10e⁻⟶ 5Pb²⁺ + 2MnO₄⁻ + 10H₂O + 16H⁺ + 10e⁻

10. Cancel species that occur on each side of the equation

5PbO₂ + 2Mn² + <u>8H₂O</u> + <u>20H⁺</u> + <u>10e⁻</u> ⟶ 5Pb²⁺ + 2MnO₄⁻ + <u>10H₂O</u> + <u>16H⁺</u> + <u>10e⁻ </u>

becomes

5PbO₂ + 2Mn²⁺ + 4H⁺ ⟶ 5Pb²⁺ + 2MnO₄⁻ + 2H₂O

11. Add the missing spectator ions

5PbO₂ + 2Mn²⁺    + 4H⁺                              ⟶            5Pb²⁺   + 2MnO₄⁻ + 2H₂O

            + 2SO₄²⁻ + 4NO₃⁻ + 2H⁺ + 2NO₃⁻       +2SO₄²⁻ + 6NO₃⁻ + 2H⁺

becomes

5PbO₂ + 2MnSO₄ + 6HNO₃ ⟶ 2PbSO₄ + 3Pb(NO₃)₂ + 2HMnO₄ + 2H₂O

12. Check that all atoms are balanced.

\begin{array}{ccc}\textbf{Atom} & \textbf{On the left} & \textbf{On the right}\\\text{Pb} & 5 & 5\\\text{O} & 36 & 36\\\text{S} & 2 & 2\\\text{H} & 6 & 6\\\text{N} & 6 & 6\\\end{array}

Everything checks. The balanced equation is

5PbO₂ + 2MnSO₄ + 6HNO₃ ⟶ 2PbSO₄ + 3Pb(NO₃)₂ + 2HMnO₄ + 2H₂O

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Chemical properties of an element are primarily dependent upon Group of answer choices the electron configuration of the element
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Answer:

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Explanation:

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