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natta225 [31]
3 years ago
7

Use the unbalanced equation NH3 + O2 = NO + H2O, what is the mole ratio between NH3 and NO

Chemistry
2 answers:
vovangra [49]3 years ago
5 0

Answer: The mole ratio between NH_3 and NO is, 1 : 1

Explanation:

The given unbalanced chemical reaction is,

NH_3+O_2\rightarrow NO+H_2O

This is an unbalanced chemical reaction because in this reaction the number oxygen and hydrogen atoms are not balanced on both side.

In order to balance the chemical reaction, the coefficient 4 is put before NH_3, coefficient 3 is put before O_2, coefficient 4 is put before NO and coefficient 2 is put before H_2O.

The balanced chemical reaction will be,

4NH_3+3O_2\rightarrow 4NO+2H_2O

From the balanced chemical reaction we conclude that,

The mole ratio of NH_3:O_2:NO:H_2O are 4 : 3 : 4 : 2

The mole ratio of NH_3:NO are 4 : 4 or 1 : 1

Therefore, the mole ratio between NH_3 and NO is, 1 : 1

kirill115 [55]3 years ago
4 0
<span>The first thing that needs to be done in order to answer the question above is to balance first the chemical equation by seeing to that the number of moles of a certain element on the reactant side is equal to the number of moles in the product side. 4NH3 + 5O2  4NO + 6H2O The mole fraction between the NH3 and NO is therefore 4:4 or 1:1.</span>
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Use the balanced equation from number four above to determine the limiting reactant if we had a mixture of 5.0 moles of Fe and 4
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Answer:

2Fe + 3O₂  →  2Fe₂O₃

Limiting reactant: O₂

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Limiting reactant: Fe

Explanation:

2Fe + 3O₂  →  2Fe₂O₃

2Fe + O₂ → 2FeO

These are the possible reactions:

In first case, 2 moles of Fe need 3 mol of oyxgen to react

If I have 5 moles of Fe, I will need (5 .3)/2 = 7.5 moles of O₂

Then, the oxygen is my limiting ( I only have 4 moles)

3 moles of O₂ need 2 moles of Fe

If I have 4 moles of O₂, I will need (4 .2)/3 = 2.66 moles (I have 5)

Fe, is the reactant in excess.

For second case, 2 moles of Fe need 1 mol of O₂, to react.

If I have 5 moles of Fe, I will need ( 5 .1) / 2 =2.5 moles of O₂

I have 4 moles of oxygen, so now it is my excess.

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Calculate the pressure in atm of .68 mol of H at 298K and occupying 4.5 L
inessss [21]

Answer:

3.7 atm

General Formulas and Concepts:

<u>Atomic Structure</u>

  • Moles

<u>Gas Laws</u>

Ideal Gas Law: PV = nRT

  • <em>P</em> is pressure
  • <em>V</em> is volume
  • <em>n</em> is number of moles
  • <em>R</em> is gas constant
  • <em>T</em> is temperature

Explanation:

<u>Step 1: Define</u>

<em>Identify variables</em>

[Given] <em>n</em> = 0.68 mol H

[Given] <em>T</em> = 298 K

[Given] <em>V</em> = 4.5 L

[Given] <em>R</em> = 0.0821 L · atm · mol⁻¹ · K⁻¹

[Solve] <em>P</em>

<em />

<u>Step 2: Find Pressure</u>

  1. Substitute in variables [Ideal Gas Law]:                                                          P(4.5 L) = (0.68 mol)(0.0821 L · atm · mol⁻¹ · K⁻¹)(298 K)
  2. Multiply [Cancel out units]:                                                                               P(4.5 L) = (0.055828 L · atm · K⁻¹)(298 K)
  3. Multiply [Cancel out units]:                                                                               P(4.5 L) = 16.6367 L · atm
  4. Isolate <em>P</em> [Cancel out units]:                                                                             P = 3.69705 atm

<u>Step 3: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs as our lowest.</em>

3.69705 atm ≈ 3.7 atm

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3 years ago
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