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Rufina [12.5K]
3 years ago
8

Nitrogen and hydrogen gases are combined at high temperatures and pressures to produce ammonia, NH3. If 101.7 g of N2 are reacte

d with excess H2, how many moles of NH3 will be formed
Chemistry
2 answers:
miskamm [114]3 years ago
7 0

Answer:

In this reaction, 7.26 moles of NH3 will be formed.

Explanation:

Step 1: Data given

Mass of N2 = 101.7 grams

Molar mass N2 = 28.0 g/mol

H2 is in excess

Molar mass H2 = 2.02 g/mol

Molar mass of NH3 = 17.03 g/mol

Step 2: The balanced equation

N2(g) + 3H2(g) → 2NH3(g)

Step 3: Calculate moles N2

Moles N2 = mass N2/ molar mass N2

Moles N2 = 101.7 grams / 28.0 g/mol

Moles N2 = 3.63 moles

Step 4: Calculate moles NH3

For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3

For 3.63 moles N2 we'll produce2*3.63 = 7.26 moles NH3

In this reaction, 7.26 moles of NH3 will be formed.

Lerok [7]3 years ago
6 0

Answer:

7.26 moles of NH₃ are formed in this reaction

Explanation:

This is about the reaction for the production of ammonia

1 mol of nitrogen gas reacts to 3 moles of hydrogen in order to produce 2 moles of ammonia.

The equation is: N₂ + 3H₂ → 2NH₃

In the question, we were informed that the excess is the H₂ so the N₂ is limiting reagent. We determine the moles, that has reacted:

101.7 g / 28 g/mol = 3.63 moles

So, If 1 mol of nitrogen gas can produce 2 moles of ammonia

3.63 moles of N₂ must produce ( 2 . 3.63) / 1 = 7.26 moles of NH₃

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2.56 g of hydrogen reacts completely with 20.32 g of oxygen<br> to form X g of water. X = g
Brilliant_brown [7]

Answer:

Mass of water produced is 22.86 g.

Explanation:

Given data:

Mass of hydrogen = 2.56 g

Mass of oxygen = 20.32 g

Mass of water = ?

Solution:

Chemical equation:

2H₂ + O₂   →  2H₂O

Number of moles of oxygen:

Number of moles = mass/ molar mass

Number of moles = 20.32 g/ 32 g/mol

Number of moles = 0.635 mol

Number of moles of hydrogen:

Number of moles = mass/ molar mass

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Number of moles = 1.28 mol

Now we will compare the moles of water with oxygen and hydrogen.

                    O₂            :            H₂O

                     1              :             2

                  0.635        ;            2×0.635 =  1.27

                   H₂             :              H₂O

                    2              :              2

                 1.28            :           1.28

The number of  moles of water produced by oxygen are less thus it will be limiting reactant.

Mass of water produced:

Mass = number of moles × molar mass

Mass = 1.27 × 18 g/mol

Mass = 22.86 g

 

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