hessaalneyadi07
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please help :)

4NH_3+ 〖3 O〗_2 → 2 NO+6 H_2 O

1. How many grams of NO can be produced from 12 g of NH3 and 12 g of O2?

2. What is the limiting reactant? What is the excess reactant?

3. How much excess reactant remains when the reaction is over?

Respuesta :

[tex]\\ \tt\hookrightarrow 4NH_3+3O_2\longrightarrow 2NO+6H_2O[/tex]

#1

We can solve through ammonia or oxygen .Lets go through ammonia

  • 4mol of ammonia produces 2mol NO
  • 2mol of ammonia produces 1mol NO
  • 1mol of ammonia produces 0.5mol NO.

Moles of Ammonia

[tex]\\ \tt\hookrightarrow \dfrac{12}{17}=0.7mol[/tex]

Moles of NO

[tex]\\ \tt\hookrightarrow 0.7(0.5)=0.35mol[/tex]

Mass of NO

  • 0.35(30)=10.5g

#2

NH_3 is excess reagent and O_2 is limiting reagent .

#3

We need ∆n

  • ∆n=8-7=1mol

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