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aev [14]
3 years ago
11

Nitrogen forms more oxides than any other element. The percents by mass of in three different nitrogen oxides are (1) (II) and (

III) 25.94 For each compound, determine (a) the simplest whole-number ratio of to and (b) the number of grams of oxygen per 1.00 of nitrogen.
Chemistry
1 answer:
Tema [17]3 years ago
3 0

Complete question;

Nitrogen forms more oxides than any other element. The percents by mass of N in three different nitrogen oxides are (|) 46.69%;(II) 36.85 %; (III) 25.94%. For each compound, determine (a) the simplest whole-number ratio of N to O, and (b) the number of grams of oxygen per 1.00 g of nitrogen.

Answer:

a. (i) The ratio is 1:1 , the formula = NO  (ii)The ratio is 1 : 1.5 which is  2 : 3, the formula = N₂O₃  (iii) The ratio is 1 : 2.5 which is 2:5 , the formula = N₂O₅

b. (i)number of grams of oxygen = 53.31/46.69 = 1.14 g

(ii)number of grams of oxygen = 63.15/36.8 = 1.71 g

(iii)number of grams of oxygen = 74.06/25.94 = 2.855 g

Explanation:

a.

(i) The percentage by mass of the nitrogen in Nitrogen oxide (i) is 46.69% which is taken as 46.69 grams . Since the other element is oxygen the mass of oxygen will be 100 - 46.69 = 53.31 grams.

The relative atomic mass of Nitrogen and oxygen is 14 amu and 16 amu respectively.

Therefore, to know the whole number ratio of N and O we find the number of moles.

number of moles of N = 46.69/14 = 3.335

number of moles of O = 53.31/16 = 3.332

The ratio is 1:1 , the formula = NO

(ii)

number of moles of N = 36.85/14 = 2.632

number of moles of O = 63.15/16 = 3.947

The ratio is 1 : 1.5 which is  2 : 3, the formula = N₂O₃

(iii)

number of moles of N = 25.94/14 = 1.85

number of moles of O = 74.06/16 = 4.63

The ratio is 1 : 2.5 which is 2:5 , the formula = N₂O₅

b.

(i) 46.69 g of nitrogen  = 53.31 g of oxygen

1 g of nitrogen = ? of Oxygen

number of grams of oxygen = 53.31/46.69 = 1.14 g

(ii)

Using similar method in b(i)

number of grams of oxygen = 63.15/36.8 = 1.71 g

(iii)

Using similar method in b(i)

number of grams of oxygen = 74.06/25.94 = 2.855 g

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The balance chemical equation is as follow,

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First find out the limiting reactant;
According to equation,

                     28 g (1 mole) N₂ reacts with  =  6 g (3 mole) of H₂
So,
                             100 g N₂ will react with  =  X g of H₂

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So, in order to consume 100 g of N₂ we will require 21.42 g of H₂, but we are provided with only 20 g of H₂, Therefore, H₂ is limiting reagent and will control the yield of product. Hence,
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                                 6 g H₂ produced  =  34 g of NH₃
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Answer:

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Explanation:

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N₂(g) + 3 H₂(g) → 2 NH₃(g)

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MW(N₂) = 2 x 14 g/mol = 28 g/mol

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MW(H₂) = 2 x 1 g/mol = 2 g/mol

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MW(NH₃) = 14 g/mol + (3 x 1 g/mol) = 17 g/mol

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Now, we have to figure out which is the limiting reactant. For this, we know that the stoichiometric ratio is 28 g N₂/6 g H₂. If we have 36.85 g of H₂, we need the following mass of N₂:

36.85 g H₂ x 28 g N₂/6 g H₂ = 171.97 g N₂

We have 23.15 g N₂ and we need 171.97 g. So, we have lesser N₂ than we need. Thus, the limiting reactant is N₂.

Now, we calculate the product (NH₃) by using the stoichiometric ratio 34 g NH₃/28 g N₂, with the mass of N₂ we have:

23.25 g N₂ x 34 g NH₃/28 g N₂ = 28.23 g NH₃

Therefore, the maximum amount of NH₃ that can be produced is 28.23 grams.

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