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NARA [144]
4 years ago
15

Liquid hexane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . suppose 41. g of hexane is mi

xed with 150. g of oxygen. calculate the maximum mass of water that could be produced by the chemical reaction. round your answer to significant digits.
Chemistry
1 answer:
Elden [556K]4 years ago
7 0
The balanced chemical equation for the reaction of hexane with O₂ is given below;
2C₆H₁₄  + 19O₂ --> 12CO₂ + 14H₂O
Stoichiometry of hexane to oxygen is 2:19
number of hexane moles - 41 g / 86.2 g/mol = 0.476 mol
number of oxygen moles - 150 g / 32 g/mol = 4.69 mol
we need to find which reagent is the limiting reactant 
If hexane is the limiting reactant 
2 moles of hexane reacts with -19 mol of oxygen 
therefore 0.476 mol - 19/2 x 0.476 = 4.522 mol
since 4.69 mol of oxygen is present, O₂ is in excess and hexane is the limiting reactant.
stoichiometry of hexane to water is 2:14
2 mol of hexane gives 14 mol of water
therefore 0.476 mol of hexane forms - 14/2 x 0.476 - 3.33 mol
mass of water formed - 3.33 mol x 18.0 g/mol = 59.9 g
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Answer:

Ka = 1.39x10⁻⁶

Explanation:

A monoprotic acid, HX, will be in equilibrium in an aqueous medium such as:

HX(aq) ⇄ H⁺(aq) + X⁻(aq)

<em>Where Ka is:</em>

Ka = [H⁺] [X⁻] / [HX]

<em>Where [] is the molar concentration in equilibrium of each specie. </em>

The equilibrium is reached when some HX reacts producing H+ and X-, that is:

[HX] = 1.64M - X

[H⁺] = X

[X⁻] = X

As pH is 2.82 = -log [H⁺]:

[H⁺] = 1.51x10⁻³M:

[HX] = 1.64M - 1.51x10⁻³M = 1.638M

[H⁺] = 1.51x10⁻³M

[X⁻] = 1.51x10⁻³M

And Ka is:

Ka = [1.51x10⁻³M] [1.51x10⁻³M] / [1.638M]

<h3>Ka = 1.39x10⁻⁶</h3>
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Answer:

1255.4L

Explanation:

Given parameters:

P₁  = 928kpa

T₁  = 129°C

V₁  = 569L

P₂ = 319kpa

T₂  = 32°C

Unknown:

V₂  = ?

Solution:

The combined gas law application to this problem can help us solve it. It is mathematically expressed as;

           \frac{P_{1} V_{1} }{T_{1} }   = \frac{P_{2} V_{2} }{T_{2} }

P, V and T are pressure, volume and temperature

where 1 and 2 are initial and final states.

Now,

 take the units to the appropriate ones;

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P₂ = 319kpa in atm gives 3.15atm

P₁  = 928kpa gives 9.16atm

T₂  = 32°C gives 273 + 32  = 305K

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Input the values in the equation and solve for V₂;

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       V₂   = 1255.4L

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