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Elza [17]
3 years ago
7

As we saw in section 4.3, nitric acid and hydrochloric acid are strong acids. write net ionic equations for the neutralization o

f the nitric acid.
Chemistry
1 answer:
Kazeer [188]3 years ago
6 0

First you write the "molecular" equation, then the ionic equation, and finally the net ionic equation.

Explanation:

You can read about these equations at

Here are the steps for your reaction.

1. Write the molecular equation

We write the equation as if all the reactants were molecules.

2HNO₃(aq) + Mg(OH)₂(s) → 2H₂O(l) + Mg(NO₃)₂(aq

2. Write the ionic equation

We write all the soluble strong electrolytes as ions. H₂O is a weak electrolyte, so we write it as a molecule. Mg(OH)₂ is insoluble, so we write it as a molecule as well.

2H⁺(aq) + 2NO₃⁻(aq) + Mg(OH)₂(s) → 2H₂O(l) + Mg²⁺(aq) + 2NO₃⁻(aq)

3. Write the net ionic equation

Here, we cancel all species that appear on both sides of the equation. We cancel the NO₃⁻ ions.

The net ionic equation is

2H⁺(aq) + Mg(OH)₂(s) → 2H₂O(l) + Mg²⁺(aq)

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In a parallel circuit, there is only one path for current to take.<br> true or false
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How many moles of CF4 can be produced when 8.95 mol C reacts with 7.88 mol F2?
Molodets [167]

The remaining moles of C is 5.01 moles while the remaining moles of F₂ is 0.

<h3>Reaction between Carbon and Fluorine </h3>

The reaction between carbon and Fluorine is given as;

C + 2F₂  -------> CF₄

1  :    2                1

from the reaction above,

2 moles of F₂ requires 1 mole of C

7.88 mole of  F₂ will require:    7.88/2 = 3.94 moles of C and 3.94 moles of CF₄.

The remaining moles of C = 8.95 - 3.94 = 5.01 moles while the remaining moles of F₂ is 0.

Learn more about moles here: brainly.com/question/15356425

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8 0
2 years ago
How many milliliters of 0.200 m fecl3 are needed to react with an excess of na2s to produce 1.38 g of fe2s3 if the percent yield
ahrayia [7]
<span>Answer: 100 ml
</span>

<span>Explanation:


1) Convert 1.38 g of Fe₂S₃ into number of moles, n


</span>i) Formula: n = mass in grass / molar mass
<span>
ii) molar mass of </span><span>Fe₂S₃ =2 x 55.8 g/mol + 3 x 32.1 g/mol = 207.9 g/mol
</span>

iii) n = 1.38 g / 207.9 g/mol = 0.00664 moles of <span>Fe₂S₃
</span>

<span>2) Use the percent yield to calculate the theoretical amount:
</span>

<span>65% = 0.65 = actual yield/ theoretical yield =>


</span>theoretical yield = actual yield / 0.65 = 0.00664 moles / 0.65 = 0.010 mol <span>Fe₂S₃</span><span>

3) Chemical equation:
</span>

<span> 3 Na₂S(aq) + 2 FeCl₃(aq) → Fe₂S₃(s) + 6 NaCl(aq)


4) Stoichiometrical mole ratios:
</span>

<span>3 mol Na₂S : 2 mol FeCl₃ : 1 mol Fe₂S₃ : 6 mol NaCl


5) Proportionality:


</span>2moles FeCl₃ / 1 mol Fe₂S₃ = x / 0.010 mol Fe₂S₃
<span>
=> x = 0.020 mol FeCl₃


6) convert 0.020 mol to volume
</span>

<span>i) Molarity formula: M = n / V
</span>

<span>ii) V = n / M = 0.020 mol / 0.2 M = 0.1 liter = 100 ml
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3 0
3 years ago
Read 2 more answers
Which reactant will be used up first if 78.1g of o2 is reacted with 62.4g of c4h10?
dlinn [17]

Answer:

Reagent O₂ will be consumed first.

Explanation:

The balanced reaction between O₂ and C₄H₁₀ is:

2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O

Then, by reaction stoichiometry, the following amounts of reactants and products participate in the reaction:

  • C₄H₁₀: 2 moles
  • O₂: 13 moles
  • CO₂: 8 moles
  • H₂O: 10 moles

Being:

  • C: 12 g/mole
  • H: 1 g/mole
  • O: 16 g/mole

The molar mass of the compounds that participate in the reaction is:

  • C₄H₁₀: 4*12 g/mole + 10*1 g/mole= 58 g/mole
  • O₂: 2*16 g/mole= 32 g/mole
  • CO₂: 12 g/mole + 2*16 g/mole= 44 g/mole
  • H₂O: 2*1 g/mole + 16 g/mole= 18 g/mole

Then, by reaction stoichiometry, the following mass quantities of reactants and products participate in the reaction:

  • C₄H₁₀: 2 moles* 58 g/mole= 116 g
  • O₂: 13 moles* 32 g/mole= 416 g
  • CO₂: 8 moles* 44 g/mole= 352 g
  • H₂O: 10 moles* 18 g/mole= 180 g

If 78.1 g of O₂ react, it is possible to apply the following rule of three: if by stoichiometry 416 g of O₂ react with 116 g of C₄H₁₀, 62.4 g of C₄H₁₀ with how much mass of O₂ do they react?

mass of O_{2} =\frac{416grams of O_{2}*62.4 grams ofC_{4}H_{10}   }{116 grams of C_{4}H_{10}}

mass of O₂= 223.78 grams

But 21.78 grams of O₂ are not available, 78.1 grams are available. Since you have less mass than you need to react with 62.4 g of C₄H₁₀, <u><em>reagent O₂ will be consumed first.</em></u>

3 0
3 years ago
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