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Limiting Reagent or Reactant

C 2 H 4 g+3 O...

Question

$C_{2} H_{4} (g) + 3 O_{2} (g) ⟶ 2 {C O}_{2} (g) + 2 H_{2} O (l)$
When $100$ grams of $O_{2}$ are allowed to react completely with $1.0$ mole of $C_{2} H_{4}$ according to the equation above, which of the following results?

Some

C_{2} H_{4}

remains unreacted

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Some

O_{2}

remains unreacted

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Only

{C O}_{2}

and

H_{2} O

are present when the reaction has run to completion

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Less than

2

moles of

{C O}_{2}

is formed

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The partial pressure of

O_{2}

falls to zero

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Solution

The correct option is B Some $O_{2}$ remains unreacted
When 100 grams of $O_{2}$ are allowed to react completely with 1.0 mole of $C_{2} H_{4}$ according to the equation $C_{2} H_{4} (g) + 3 O_{2} (g) ⟶ 2 {C O}_{2} (g) + 2 H_{2} O (l)$ , some $O_{2}$ remains unreacted.
The molecular weight of $O_{2}$ is 32 g/mol.
100 grams of $O_{2}$ corresponds to $\frac{100}{32} = 3.125$ moles.
According to the given equation ,
1.0 mole of $C_{2} H_{4}$ reacts with 3 moles of $O_{2}$ . Hence $C_{2} H_{4}$ is limiting reagent and $O_{2}$ is excess reagent. Hence, some $O_{2}$ remains unreacted.

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and

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Q. Calculate the heat of hydrogenation of

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C_{2} H_{4}

H_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ H_{2} O (l); Δ H = - 68.32 k c a l

C_{2} H_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ 2 C O_{2} (g) + H_{2} O (l); Δ H = - 310.61 k c a l

C_{2} H_{4} (g) + 3 O_{2} (g) ⟶ 2 C O_{2} (g) + H_{2} O (l); Δ H = - 337.32 k c a l

20 m l

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Q. Complete and balance reaction of the given reaction is:

{C r}_{2} O_{7}^{2 -} + C_{2} H_{4} O ⟶ C_{2} H_{4} O_{2} + {C r}^{3 +}

Q. Write correctly balanced half reaction and overall equation for the following skeletal equation :
A)

{N O}_{3}^{-} + B i (s) \to {B i}^{3 +} + {N O}_{2}

(in acid solution)
B)

F e {(O H)}_{3} (s) + H_{2} O_{2} \to F e {(O H)}_{2} (s) + H_{2} O + O_{2}

(in basic medium)
C)

{C r}_{2} O_{7}^{2 -} + C_{2} H_{4} O \to C_{2} H_{4} O_{2} + {C r}^{3 +}

(in acid solution)

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C2H4(g)+3O2(g)⟶2CO2(g)+2H2O(l)When 100 grams of O2 are allowed to react completely with 1.0 mole of C2H4 according to the equation above, which of the following results?

$C_{2} H_{4} (g) + 3 O_{2} (g) ⟶ 2 {C O}_{2} (g) + 2 H_{2} O (l)$
When $100$ grams of $O_{2}$ are allowed to react completely with $1.0$ mole of $C_{2} H_{4}$ according to the equation above, which of the following results?