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nadya68 [22]
4 years ago
14

Mass of oxygen gas needed to burn 37.5g of wax

Chemistry
1 answer:
Slav-nsk [51]4 years ago
6 0
Answer is: mass oxygen is 129,28 grams.
Chemical reaction: 2C₁₈H₃₈ + 55O₂ → 36CO₂ + 38H₂O.
m(C₁₈H₃₈) = 37,5 g.
n(C₁₈H₃₈) = m(C₁₈H₃₈) ÷ M(C₁₈H₃₈).
n(C₁₈H₃₈) = 37,5 g ÷ 254 g/mol.
n(C₁₈H₃₈) = 0,147 mol.
From chemical reaction: n(C₁₈H₃₈) : n(O₂) = 2 : 55.
n(O₂) = 4,04 mol.
m(O₂) = 4,04 mol · 32 g/mol.
m(O₂) = 129,28 g.
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The concentration of CO is 4.33*10⁻³\frac{moles}{L}

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A reversible chemical reaction, as in this case, occurs in both directions: reagents transforming into products (direct reaction) and products transforming back into reagents (reverse reaction).

Chemical Equilibrium is the state in which the direct and indirect reaction have the same reaction rate and is defined by a chemical constant Kc.

Being:

aA + bB ⇔ cC + dD

where A, B, C and D represent the chemical species involved and a, b, c and d their respective stoichiometric coefficients, constant Kc is defined as:

Kc=\frac{[C]^{c}*[D]^{d}  }{[A]^{a} *[B]^{b} }

In the case of the reaction

2 H₂(g) + CO(g) ⇌ CH₃OH(g)

the equilibrium constant Kc is:

Kc=\frac{[CH_{3}OH] }{[H_{2} ]^{2}*[CO] }

You know:

  • Kc=35
  • [CH₃OH] =\frac{4.87*10^{-3}moles }{3.63 L} =1.34*10^{-3} \frac{moles}{L}
  • [H₂]=\frac{3.21*10^{-2}moles }{3.63 L} =8.84*10^{-3} \frac{moles}{L}
  • [CO]=?

Replacing:

35=\frac{1.34*10^{-3} }{(8.84*10^{-3} )^{2} *[CO]}

Solving you get:

[CO]= \frac{1.34*10^{-3} }{(8.84*10^{-3} )^{2} *35}

[CO]=4.33*10⁻³\frac{moles}{L}

<u><em>The concentration of CO is 4.33*10⁻³</em></u>\frac{moles}{L}<u><em /></u>

8 0
4 years ago
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