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Ne4ueva [31]
3 years ago
6

The combustion of ammonia in the presence of excess oxygen yields no2 and h2o: 4 nh3 (g) + 7 o2 (g) â 4 no2 (g) + 6 h2o (g) the

combustion of 28.8 g of ammonia consumes ________ g of oxygen.
Chemistry
1 answer:
VLD [36.1K]3 years ago
8 0
The answer would be 118.68 g.
Explanation for this is:4 moles of NH3 give 4 moles of NO2 
so 1mole of NH3 will give 1 mole of NO2 
43.9 grams of NH3 contains 2.58 moles 
so 2.58 moles will be produced of NO2 
which is 118.7 grams this the amount of oxygen that is used.
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In a certain experiment, 28.0 mL of 0.250 M HNO3and 53.0 mL of 0.320 M KOH are mixed. Calculate the number of moles of water for
Xelga [282]

Answer:

The number of moles of water formed in the resulting reaction is 6.03

[H+]: 37,2 M

[OH-]: 37,2 M

Explanation:

HNO3  +  KOH ----> KNO3 + H2O

First, we must discover the limiting reagent and we need to find out the moles, we use for this.

Moles that are used = Molarity / volume

HNO3 : 0,250 mol/L / 0,028L = 8,93 moles

KOH : 0,320 mol/L / 0,053L = 6,03 moles

The ratio of the reagents by stoichiometry is 1 to 1, so the limiting reagent is KOH, if I need 1 mole of nitric per mole of KOH, for every 8.93 moles I will need the same. However I have only 6.03 moles of KOH

The ratio of the reagents/products by stoichiometry is 1 to 1 so if I need 1 mol of KOH to make 1 mol of Water, 6,03 moles of KOH are used to make 6,03 moles of H2O.

The equilibrium of water is this:

2H2O ⇄ H3O+  +  OH-

2 moles of water are broken down into 1 mole of hydronium (H3O +) and 1 mole of hydroxyl (OH-)

6,03 moles of water are broken down into the half of those moles, so we have 3,015 moles of H3O+ and 3,015 moles of OH- but these moles are in 81,0 mL (the volume of the two solutions, 28 mL + 53 mL)

We must find out the moles in 1000 mL (1 L) so let's apply the rule of three.

81 mL ____ 3,015 moles

1000 mL ___ ( 1000 . 3,015) /81 = 37,2 M

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