Name the type of reaction shown:2C4H10+13O2=8CO2+10H2O

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

Feliz [49]

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

2Mn²⁺ + 8H₂O ⟶ 2MnO₄⁻ + 16H⁺ + 10e⁻

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

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

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

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

7 0

4 years ago

Cal is titrating 50.8 mL of 0.319 M HBr with 0.337 M Ba(OH)2. How many mL of Ba(OH)2 does Cal need to add to reach the equivalen

alexgriva [62]

Answer:

$V_{base}=24.04mL$

Explanation:

Hello!

In this case, according to the chemical reaction by which HBr reacts with Ba(OH)2:

$2HBr+Ba(OH)_2\rightarrow BaBr_2+2H_2O$

We can see there is a 2:1 mole ratio between the acid and the base; thus, at the equivalent point we can write:

$2M_{base}V_{base}=M_{acid}V_{acid}$

Therefore, for is to compute the volume of the used base, we proceed as shown below:

$V_{base}=\frac{M_{acid}V_{acid}}{2M_{base}}$

And we plug in to obtain:

$V_{base}=\frac{0.319M*50.8mL}{2*0.337M}\\\\V_{base}=24.04mL$

Best regards!

8 0

3 years ago

Help on this question please

oee [108]

Answer:

First choice: 2

Explanation:

There are 2 phosphorous (P) in the substance.

Ignore the strontium (Sr3) part because you are looking to isolate the P from (PO4)2.

Break the chemical equation apart to get 1 Phosphorous atom, and 4 Oxygen atoms.

Now, multiple 1 by 2 because that are 2 phosphate to get 2 phosphorous atoms.

8 0

3 years ago

What shows an atom correctly?

castortr0y [4]

Answer:

B

Explanation:

Atom consists of proton and neutron in center (often referred as nucleus) and electrons circling those things.

5 0

3 years ago

Read 2 more answers

Answer the following question attached!!

irina [24]

Answer:

I think It is Covalent. Not 100% sure but like 90% sure.

4 0

3 years ago