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How many Grams of C₆H₆ must decompose according to the following chemical equation to transfer 430kj of heat?
C₆H₆(l) → 3 C₂H₂(g)∆H= 630 kj

Respuesta :

Answer:

53·32 g

Explanation:

Given the ΔH for the reaction is for 1 mole of benzene(C6H6) and 1 mole of benzene contains 78 g of benzene

So for 78 g of benzene ΔH of the reaction is 630 kJ

Let the weight of benzene that must decompose to transfer 430 kJ of heat be m g

∴     78 ≡ 630

      m ≡ 430

For 1 kJ of heat to be transferred the weight of benzene required is 78 ÷ 630 = 0·124 g

For 430 kJ of heat to be transferred the weight of benzene required is 0·124 × 430 = 53·32 g

∴ m = 53·32 g

∴ 53·32 grams of C6H6 must decompose according to the given chemical equation to transfer 430 kJ of heat

Answer:

THE NUMBER OF GRAMS OF [tex]C_6H_6[/tex] PRODUCED IS [tex]58.23812 g[/tex]

Explanation:

The given equation is ,

  • [tex]C_6H_6(l)[/tex]⇒[tex]3C_2H_2(g)[/tex]

ΔH=[tex]630kJ[/tex];

  • Thus , for one mole of [tex]C_6H_6[/tex] we will require [tex]630kJ[/tex] of energy to completely convert it into 3 moles of [tex]3C_2H_2[/tex].
  • Thus by giving , [tex]430kJ[/tex] of energy ,

the number of moles '[tex]n[/tex]' that will decompose is :

[tex]n=\frac{430}{630}  \\\\n=\frac{43}{63} \\n=0.68254[/tex]

Thus,

The number of grams [tex]w[/tex] = no.of moles [tex]*[/tex] weight of one mole

weight of one mole = 6[tex]*[/tex](mass of one mole of [tex]C[/tex]) + 6[tex]*[/tex](mass of one mole of [tex]H[/tex])

  • weight of one mole = [tex]6*12+6*1\\=72+6\\=78[/tex];
  • [tex]w=0.68254*(78)\\w=58.23812 g[/tex]

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