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mezya [45]
3 years ago
13

Consider the following balanced reaction between hydrogen and nitrogen to form ammonia: 3H2(g) + N2(g)→2NH3(g)

Chemistry
1 answer:
andrew-mc [135]3 years ago
8 0

the balanced equation for the formation of ammonia is as follows

3H₂ + N₂ --> 2NH₃

we are told that N₂ is in excess. this means that H₂ is the limiting reactant

limiting reactant is fully consumed in the reaction , amount of product formed depends on amount of limiting reactant present

stoichiometry of H₂ to NH₃ is 3:2

when 3 mol of H₂ reacts - 2 mol of NH₃ are formed

therefore when 24.0 mol of H₂ reacts - 2/3 x 24.0 = 16 mol of NH₃ are formed

16 mol of NH₃ are formed

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Calculate the empirical formula for each stimulant based on its elemental mass percent composition.
Vanyuwa [196]
A.   
elements                              C                H          N
percentage composition    74.03          8.70       17.27
Molecular mass                   12                1          14

# of mole                            6.17              8.70        1.23

÷smallest mole                    5.0                 7.0           1.0

mole ratio                              5    :               7      ;       1

THE EMPERICAL FORMUKLA FOR A. IS  C5H7O

NOTE: #of mole = percentage composition ÷ Mr
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and b. should be done using the same procedure
3 0
3 years ago
Read 2 more answers
What is the enthalpy of the overall chemical reaction ch4(g)+4cl2(g)- ccl4(g)+4hcl(g)?
Brut [27]

Answer:

-205.7kj

Explanation:

Now adding reaction 2 and twice of reaction 3 and reverse of reaction 1, we get the enthalpy of the reaction.

The expression for enthalpy for the following reaction will be,

where,

n = number of moles

Now put all the given values in the above expression, we get:

Therefore, the enthalpy of the following reaction is,  -205.7kj

4 0
3 years ago
Are the following quantum number combinations allowed? If not, show two ways to correct them:
evablogger [386]

Answer:

(a) n = 2; l = 0; ml = -1 ⇒ WRONG

change ml = 0, or l = 1 to correct.

(b) n = 4, l = 3; ml = -1 ⇒ CORRECT

(c) n = 3, l = 1; ml = 0 ⇒ CORRECT

(d) n = 5, l = 2; ml = +3 ⇒ WRONG

change ml = 0, ±1 or ±2, or l = 3 or l = 4 to correct.

Explanation:

The rules for electron quantum numbers are:

1. Shell number, 1 ≤ n

2. Subshell number, 0 ≤ l ≤ n − 1

3. Orbital energy shift, -l ≤ ml ≤ l

4. Spin, either -1/2 or +1/2

Now, apply in our case

(a) n = 2; l = 0; ml = -1 ⇒ WRONG

-l ≤ ml ≤ l, so either change ml = 0 or l = 1 to correct.

(b) n = 4, l = 3; ml = -1 ⇒ CORRECT

(c) n = 3, l = 1; ml = 0 ⇒ CORRECT

(d) n = 5, l = 2; ml = +3 ⇒ WRONG

-l ≤ ml ≤ l, so either change ml = 0, ±1 or ±2, or l = 3 or l = 4 to correct.

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3 years ago
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Answer:

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Answer: C.)

Explanation:

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