The complete balanced chemical
equation is:
4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (g)
In statement form: 4mol NH3 reacts with 5 mol O2 to produce 6
mol H2O
First let us find for the limiting reactant:
>molar mass NH3 = 17 g/mol
moles NH3 = 54/17 = 3.18 mol NH3
This will react with 3.18*5/4 = 3.97 mol O2
>molar mass O2 = 32g/mol
moles O2 = 54/32 = 1.69 mol O2
We have insufficient O2 therefore this is the limiting
reactant
From the balanced equation:
For every 5.0 mol O2, we get 6.0 mol H2O, therefore
moles H2O formed = 1.69
mol O2 * 6/5 = 2.025 mol
Molar mass H2O = 18g/mol
<span>mass H2O formed = 2.025*18 = 36.45 grams H2O produced</span>
Answer:
The answer to your question is it is not at equilibrium, it will move to the products.
Explanation:
Data
Keq = 2400
Volume = 1 L
moles of NO = 0.024
moles of N₂ = 2
moles of O₂ = 2.6
Process
1.- Determine the concentration of reactants and products
[NO] = 0.024 / 1 = 0.024
[N₂] = 2/1 = 2
[O₂] = 2.6/ 1= 2.6
2.- Balanced chemical reaction
N₂ + O₂ ⇒ 2NO
3.- Write the equation for the equilibrium of this reaction
Keq = [NO]²/[N₂][O₂]
- Substitution
Keq = [0.024]² / [2][2.6]
-Simplification
Keq = 0.000576 / 5.2
-Result
Keq = 1.11 x 10⁻⁴
Conclusion
It is not at equilibrium, it will move to the products because the experimental Keq was lower than the Keq theoretical-
1.11 x 10⁻⁴ < 2400
Answer:
Explanation:
C = 49.48
H = 5.19
O = 16.48
N = 28.85
ratio of moles
= 49.48 / 12 : 5.19 / 1 : 16.48 / 16 : 28.85 / 14
= 4.123 : 5.19 : 1.03 : 2.06
= 4 : 5 : 1 : 2
so the empirical formula = C₄ H₅O N₂
Let molecular formula = ( C₄ H₅ON₂ )ₙ ,
n ( 48 + 5 + 16 + 28 ) = 119.19
97 n = 194.19
n = 2 ( approx )
molecular formula = C₈ H₁₀O₂ N₄
D = m/v. v = (3)^3 = 27.
D = 27/27. D = 1g/cm^3
