Question

The ostwald process is used commercially to produce nitric acid, which is, in turn, used in many modern chemical processes. in the first step of the ostwald process, ammonia is reacted with oxygen gas to produce nitric oxide and water. what is the maximum mass of h2o that can be produced by combining 54.0 g of each reactant?

Answer 1

The complete balanced chemical equation is: 
4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (g) 

In statement form: 4mol NH3 reacts with 5 mol O2 to produce 6 mol H2O 

First let us find for the limiting reactant: 
>molar mass NH3 = 17 g/mol 
moles NH3 = 54/17 = 3.18 mol NH3 
This will react with 3.18*5/4 = 3.97 mol O2 

>molar mass O2 = 32g/mol 
moles O2 = 54/32 = 1.69 mol O2 
We have insufficient O2 therefore this is the limiting reactant 

From the balanced equation: 
For every 5.0 mol O2, we get 6.0 mol H2O, therefore

moles H2O formed =  1.69 mol O2  * 6/5 = 2.025 mol
Molar mass H2O = 18g/mol 
mass H2O formed = 2.025*18 = 36.45 grams H2O produced

Answer 2

Answer:

36.45 g

Explanation:

The balanced reaction given is:

2NH₃(g) + (5/2)O₂(g) → 2NO(g) + 3H₂O(l)

The molar masses of the interest compounds are:

NH₃ = 17.0 g/mol

O₂ = 32.0 g/mol

H₂O = 18.0 g/mol

First, let's find out which of the reactants will be totally consumed, the limiting, and which one is in excess.

By the reaction, the stoichiometry is:

2 moles of NH₃ -------------- 5/2 moles of O₂

Converting by mass (multiplying the molar mass by the number of moles), and supposing that all the ammonia will be consumed:

34 g of NH₃ ------------- 80 g of O₂

54.0 g          ------------- x

By a simple direct three rule:

34x = 4320

x = 127 g of O₂

So, it'll be needed more oxygen than we have, so, oxygen must the limiting reactant.

The stoichiometry between oxygen and water is:

5/2 moles of O₂ -------------------- 3 moles of H₂O

Converting by mass:

80 g of O₂ ------------------------ 54 g of H₂O

54 g           ------------------------ y

By a simple direct three rule:

80y = 2916

y = 36.45 g of H₂O