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andrey2020 [161]
3 years ago
10

Write a balanced chemical equation, including states of matter, for the combustion of gaseous benzene, c6h6.

Chemistry
1 answer:
snow_tiger [21]3 years ago
4 0
Almost all hydrocarbon 'burn' reactions involve oxygen; it's by far the most reactive substance in air. 

<span>Hydrocarbon combustions always involve </span>
<span>[some hydrocarbon] + oxygen --> carbon dioxide + steam. </span>

C6H6(l) + O2 (g)--> CO2 (g)+ H2O (g)

<span>Balance carbon, six on each side: </span>
C6H6(l) + O2 (g)--> 6CO2 (g)+ H2O (g)

<span>Balance hydrogen, six on each side: </span>
C6H6(l) + O2 (g)--> 6CO2(g) + 3H2O (g)

<span>Now, we have fifteen oxygens on the right and O2 on the left. </span>
<span>Two ways to deal with that. We can use a fraction: </span>
C6H6 (l)+ (15/2)O2 (g)--> 6CO2 (g)+ 3H2O (g)

<span>Or, if you prefer to have whole number coefficients, double everything </span>
<span>to get rid of the fraction: </span>
2C6H6 (l)+ 15O2 (g)--> 12CO2 (g)+ 6H2O (g)

<span>With the SATP states thrown in... </span>
C6H6(l) + (15/2)O2(g) --> 6CO2(g) + 3H2O(g)
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Answer:

11.31g NaClO₂

Explanation:

<em> Is given 250mL of a 1.60M chlorous acid HClO2 solution. Ka is 1.110x10⁻². What mass of NaClO₂ should the student dissolve in the HClO2 solution to turn it into a buffer with pH =1.45? </em>

It is possible to answer this question using Henderson-Hasselbalch equation:

pH = pKa + log₁₀ [A⁻] / [HA]

<em>Where pKa is -log Ka = 1.9547; [A⁻] is the concentration of the conjugate base (NaClO₂), [HA] the concentration of the weak acid</em>

You can change the concentration of the substance if you write the moles of the substances:

[Moles HClO₂] = 250mL = 0.25L×(1.60mol /L) = <em>0.40 moles HClO₂</em>

Replacing in H-H expression, as the pH you want is 1.45:

1.45 = 1.9547 + log₁₀ [Moles NaClO₂] / [0.40 moles HClO₂]

-0.5047 = log₁₀ [Moles NaClO₂] / [0.40 moles HClO₂]

<em>0.3128 = </em>[Moles NaClO₂] / [0.40 moles HClO₂]

0.1251 = Moles NaClO₂

As molar mass of NaClO₂ is 90.44g/mol, mass of 0.1251 moles of NaClO₂ is:

0.1251 moles NaClO₂ ₓ (90.44g / mol) =

<h3>11.31g NaClO₂</h3>
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