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Balance the following equation. Then, given the moles of reactant or product below, determine the corresponding amount in moles of each of the other reactants and products.NH3+O2⟶N2+H2O
a.4molNH3b.4 molN2c. 4.5molO2

Respuesta :

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Answer:

Option a → 4 mol NH₃

Explanation:

This the unbalanced reaction

NH₃  +  O₂  ⟶  N₂  +  H₂O

The balanced reaction:

4NH₃  +  3O₂  →  2N₂  +  6H₂O

4 mol of ammonia

3 mol of oxygen

2 mol of nitrogen

6 mol of water

Cricetus

The balanced equation will be "[tex]4NH_3 + 3O_2 \rightarrow 2N_2 + 6H_2O[/tex]". A complete explanation is below.

(a) 4 mol NH₃:

  • [tex]4 \ mol \ NH_3 (\frac{3 \ mol \ O_2}{4 \ mol \ NH_3} ) = 3 \ mol \ O_2[/tex]
  • [tex]4 \ mol \ NH_3 (\frac{2 \ mol \ N_2}{4 \ mol \ NH_3} ) = 2 \ mol \ N_2[/tex]
  • [tex]4 \ mol \ NH_3 (\frac{6 \ mol \ H_2 O}{4 \ mol \ NH_3} ) = 6 \ mol \ H_2O[/tex]

(b) 4 mol N₂:

  • [tex]4 \ mol \ N_2 (\frac{4 \ mol \ NH_3}{2 \ mol \ N_2} ) = 8 \ mol \ NH_3[/tex]
  • [tex]4 \ mol \ N_2 (\frac{ 3 \ mol \ O_2}{2 \ mol \ N_2} ) = 6 \ mol \ O_2[/tex]
  • [tex]4 \ mol \ N_2(\frac{6 \ mol \ H_2O}{2 \ mol \ N_2} ) = 12 \ mol \ H_2O[/tex]

(c) 4.5 mol O₂:

  • [tex]4.5 \ mol \ O_2 (\frac{4 \ mol \ NH_3}{3 \ mol \ O_2} ) = 6 \ mol \ NH_3[/tex]
  • [tex]4.5 \ mol \ O_2 (\frac{2 \ mol \ N_2}{3 \ mol \ O_2} ) = 3 \ mol \ N_2[/tex]
  • [tex]4.5 \ mol \ O_2 (\frac{6 \ mol \ H_2 O}{3 \ mol \ O_2} ) = 9 \ mol \ H_2O[/tex]

Thus the above equation is correct.

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