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oksian1 [2.3K]
3 years ago
12

Acetylene (c2h2) undergoes combustion in excess oxygen to generate gaseous carbon dioxide and water. given δh°f[co2(g)] = –393.5

kj/mol, δh°f[h2o(g)] = –241.8 kj/mol, and δh°f[c2h2(g)] = 226.6 kj/mol, how much energy is released (kj) when 10.5 moles of acetylene is burned?
Chemistry
1 answer:
Brums [2.3K]3 years ago
8 0
Answer is: 13181,7 kJ of energy <span>is released when 10.5 moles of acetylene is burned.
</span>Balanced chemical reaction: C₂H₂ + 5/2O₂ → 2CO₂ + H₂O.
<span>ΔHrxn = sum of ΔHf (products of reaction) - sum of ΔHf (reactants).</span><span>
Or ΔHrxn = ∑ΔHf (products of reaction) - ∑ΔHf (reactants).
ΔHrxn - enthalpy change of chemical reaction.
<span>ΔHf - enthalpy of formation of reactants or products.
</span></span>ΔHrxn = (2·(-393,5) + (-241,8)) - 226,6 · kJ/mol.
ΔHrxn = -1255,4 kJ/mol.
Make proportion: 1 mol (C₂H₂) : -1255,4 kJ = 10,5 mol(C₂H₂) : Q.
Q = 13181,7 kJ.
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When 69.9 g heptane is burned it releases __ mol water.
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Answer:

1) When 69.9 g heptane is burned it releases 5.6 mol water.  

2) C₇H₁₆ + 11O₂ → 7CO₂ + 8H₂O.

Explanation:

  • Firstly, we should balance the equation of heptane combustion.
  • The balanced equation is: <em>C₇H₁₆ + 11O₂ → 7CO₂ + 8H₂O.</em>

This means that every 1.0 mole of complete combustion of heptane will release 8 moles of H₂O.

  • We need to calculate the no. of moles in 69.9 g of heptane that is burned using the relation: <em>n = mass/molar mass.</em>

n of 69.9 g of heptane = mass/molar mass = (69.9 g)/(100.21 g/mol) = 0.697 mol ≅ 0.7 mol.

<em><u>Using cross multiplication:</u></em>

1.0 mol of heptane releases → 8 moles of water.

0.7 mol of heptane releases → ??? moles of water.

<em>∴ The no. of moles of water that will be released from burning (69.9 g) of water</em> = (0.7 mol)(8.0 mol)/(1.0 mol) = <em>5.6 mol.</em>

<em>∴ When 69.9 g heptane is burned it releases </em><em>5.6</em><em> mol water. </em>

<em />

5 0
3 years ago
N each reaction box, place the best reagent or reactant from the list below. reagents may be used more than once or not at all.
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Answer:

See explaination

Explanation:

See attachment for the drawing of the intermediate products b and c (both are neutral; omit byproducts).

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How many valence electrons does the lewis structure for a chlorine atom show? the molecule structure of cl has with electrons a.
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Seven valence electrons are present in the 'Lewis structure' for the given chlorine atom

A Chlorine atom has total 17 electrons in total. In the 'first orbit' 2 electrons, 10 electrons in the 'second orbit' and finally 7 electrons in the last orbit. As Chlorine is a Group 7 element, chlorine will have 7 'valence electrons' in its outermost orbit. While drawing the 'Lewis structure' of Chlorine we shall put seven dots that is the electrons or valance electrons around the symbol of Chlorine (CI).

<h3>What are Valence electrons ?</h3>

The electrons in an atom's outermost shell, or energy level, are called valence electrons. For instance, the valence electrons of oxygen are six, with two in the 2s subshell and four in the 2p subshell.

  • An atom's outer shell electrons, known as valence electrons, can take role in the creation of chemical bonds. When two atoms establish a single covalent bond, normally both atoms contribute one valence electron to create a shared pair. The ground state of an atom is its state with the least amount of energy.

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2 years ago
An apple has a mass of 1.2 hectograms. What is the mass of the
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3 0
3 years ago
A solution is prepared by mixing 50.0 mL of 0.50 M Cu(NO3)2 with 50.0 mL of 0.50 M Co(NO3)2. Sodium hydroxide is added to the mi
vekshin1

Answer:

Cu(OH)₂ will precipitate first, with [OH⁻] = 2.97x10⁻¹⁰ M

Explanation:

The equilibriums that take place are:

Cu⁺² + 2OH⁻ ↔ Cu(OH)₂(s)   ksp = 2.2x10⁻²⁰ = [Cu⁺²]*[OH⁻]²

Co⁺² + 2OH⁻ ↔ Co(OH)₂(s)   ksp = 1.3x10⁻¹⁵ = [Co⁺²]*[OH⁻]²

Keep in mind that <em>the concentration of each ion is halved </em>because of the dilution when mixing the solutions.

For Cu⁺²:

2.2x10⁻²⁰ = [Cu⁺²]*[OH⁻]²

2.2x10⁻²⁰ = 0.25 M*[OH⁻]²

[OH⁻] = 2.97x10⁻¹⁰ M

For Co⁺²:

1.3x10⁻¹⁵ = [Co⁺²]*[OH⁻]²

1.3x10⁻¹⁵ = 0.25 M*[OH⁻]²

[OH⁻] = 7.21x10⁻⁸ M

<u>Because Copper requires less concentration of OH⁻ than Cobalt</u>, Cu(OH)₂ will precipitate first, with [OH⁻] = 2.97x10⁻¹⁰ M

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4 years ago
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