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SSSSS [86.1K]
3 years ago
9

Balance the following half reaction in basic conditions. Then, indicate the coefficients for H2O and OH– for the balanced half r

eaction, and which side they appear on. Si (s) + Mg(OH)2 (s) → Mg (s) + SiO32- (aq)
Chemistry
1 answer:
Ugo [173]3 years ago
7 0

Answer:

The ballance half reactions are:

Mg²⁺  + 2e⁻ → Mg

6OH⁻ + Si  → SiO₃²⁻ + 4e⁻ + 3 H₂O

Coefficients for H2O and OH– are 3 for H₂O (in products side) and 6 for OH⁻ (in reactants side)

Explanation:

Si (s) + Mg(OH)₂ (s) → Mg (s) + SiO₃²⁻ (aq)

Let's see the oxidations number.

As any element in ground state, we know that oxidation state is 0, so Si in reactants and Mg in products, have 0.

Mg in reactants, acts with +2, so the oxidation number has decreased.

This is the reduction, so it has gained electrons.

Si in reactants acts with 0 so in products we find it with +4. The oxidation number increased it, so this is oxidation. The element has lost electrons.

Let's take a look to half reactions:

Mg²⁺  + 2e⁻ → Mg

Si  → SiO₃²⁻ + 4e⁻

In basic medium, we have to add water, as the same amount of oxygen we have, IN THE SAME SIDE. We have 3 oxygens in products, so we add 3 H₂O and in the opposite site we can add OH⁻, to balance the hydrogen. The half reaciton will be:

6OH⁻ + Si  → SiO₃²⁻ + 4e⁻ + 3 H₂O

If we want to ballance the main reaction we have to multiply (x2) the half reaction of oxidation. So the electrons can be ballanced.

2Mg²⁺  + 4e⁻ → 2Mg

Now, that they are ballanced we can sum the half reactions:

2Mg²⁺  + 4e⁻ → 2Mg

6OH⁻ + Si  → SiO₃²⁻ + 4e⁻ + 3 H₂O

2Mg²⁺  + 4e⁻  + 6OH⁻ + Si  → 2Mg  +  SiO₃²⁻ + 4e⁻ + 3 H₂O

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andriy [413]

Answer:

5.5 L

Explanation:

First we <u>convert 10 g of propane gas</u> (C₃H₈) to moles, using its <em>molar mass</em>:

  • 10 g ÷ 44 g/mol = 0.23 mol

Then we <u>use the PV=nRT formula</u>, where:

  • P = 1 atm & T = 293 K (This are normal conditions of T and P)
  • n = 0.23 mol
  • R = 0.082 atm·L·mol⁻¹·K⁻¹
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1 atm * V = 0.23 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 293 K

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2 years ago
11. The oxyfuel flame was used for fusion welding as early as the first half of the
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The Oxyfuel gas or flame refers to a group of welding processes that use the flame produced by the combination of a fuel gas and oxygen as the source of heat.

<u>Explanation:</u>

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Fe2O3 + CO --&gt; Fe + CO2
satela [25.4K]

Answer:

The answer to your question is

1.-Fe₂O₃

2.- 280 g

3.- 330 g

Explanation:

Data

mass of CO = 224 g

mass of Fe₂O₃ = 400 g

mass of Fe = ?

mass of CO₂

Balanced chemical reaction

                       Fe₂O₃   + 3CO    ⇒  2Fe  +   3CO₂

1.- Calculate the molar mass of Fe₂O₃ and CO

Fe₂O₃ = (56 x 2) + (16 x 3) = 160 g

CO = 12 + 16 = 28 g

2.- Calculate the proportions

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experimental proportion Fe₂O₃ / CO = 400/224 = 1.78

As the experimental proportion is lower than the theoretical, we conclude that the Fe₂O₃ is the limiting reactant.

3.-     160 g of Fe₂O₃  --------------- 2(56) g of Fe

         400 g of Fe₂O₃ ---------------  x

         x = (400 x 112) / 160

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          400 g of Fe₂O₃  --------------  x

          x = (400 x 132)/160

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Answer:

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