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Stoichiometric Calculations

In the follow...

Question

In the following reaction:
$4 N H_{3} (g) + 5 O_{2} (g) \to 4 N O (g) + 6 H_{2} O (l)$
when 1 mole ammonia and 1 mole of $O_{2}$ are mixed, then the number of moles of NO formed will be:

0.8

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0.7

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0.6

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0.5

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Solution

The correct option is A 0.8
$4 N H_{3} (g) + 5 O_{2} (g) \to 4 N O (g) + 6 H_{2} O (l)$
According to the given equation:
1 mole of $N H_{3}$ (on complete reaction) gives 1 mole $N O .$
Similarly, 1 mole of $O_{2}$ (on complete reaction) gives $\frac{4}{5}$ , i.e. ,0.8 mole $N O .$
Thus, $O_{2}$ will be limiting reactant and actual amount of $N O$ formed in the reaction will be $= 0.8 m o l e$

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Similar questions

4 N H_{3} (g) + 5 O_{2} (g) \to 4 N O (g) + 6 H_{2} O (g)

1

1

O_{2}

4 {N H}_{3} (g) + 5 O_{2} (g) \to 4 N O (g) + 6 H_{2} O (g)

{N H}_{3}

2.513

N O

Δ_{f} H^{⊖}

N H_{3} (g)

N O (g)

H_{2} O (l)

- 11.04

+ 21.60

- 68.32 K c a l

4 N H_{3} (g) + 5 O_{2} (g) ⟶ 4 N O (g) + 6 H_{2} O (l)

{4 N H}_{3 (g)} + {5 O}_{2 (g)} \to {4 N O}_{(g)} + {6 H}_{2} O_{g}

{2 N O}_{g} + O_{2 (g)} \to {2 N O}_{2 (g)}

{3 N O}_{2 (g)} + H_{2} O_{(g)} \to {2 H N O}_{3 (a q)} + {N O}_{(g)}

27^{o} C

{4 N H}_{3} (g) + {5 O}_{2} (g) ⟶ 4 N O (g) + {6 H}_{2} O (l)

Δ H = 9080 J {m o l}^{- 1} a n d Δ S = 35.7 {J K}^{- 1} {m o l}^{- 1}

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