Answer:10,900,000,000,000 ozone moles
Explanation:
Answer:
The theoretical yield of
Li
3
N
is
20.9 g
.
Explanation:
Balanced Equation
6Li(s)
+
N
2
(
g
)
→
2Li
3
N(s)
In order to determine the theoretical yield, we must first find the limiting reactant (reagent), which will determine the greatest possible amount of product that can be produced.
Molar Masses
Li
:
6.941 g/mol
N
2
:
(
2
×
14.007
g/mol
)
=
28.014 g/mol
Li
3
N
:
(
3
×
6.941
g/mol Li
)
+
(
1
×
14.007
g/mol N
)
=
34.83 g/mol Li
3
N
Limiting Reactant
Divide the mass of each reactant by its molar mass, then multiply times the mole ratio from the balanced equation with the product on top and the reactant on bottom, then multiply times the molar mass of
Li
3
N
.
Lithium
12.5
g Li
×
1
mol Li
6.941
g Li
×
2
mol Li
3
N
6
mol Li
×
34.83
g Li
3
N
1
mol Li
3
N
=
20.9 g Li
3
N
Nitrogen Gas
34.1
g N
2
×
1
mol N
2
28.014
g N
2
×
2
mol Li
3
N
1
mol N
2
×
34.83
g Li
3
N
1
mol Li
3
N
=
84.8 g Li
3
N
Lithium produces less lithium nitride than nitrogen gas. Therefore, the limiting reactant is lithium, and the theoretical yield of lithium nitride is
20.9 g
.
Explanation:
Answer:
1.8 x 10⁻⁵
Explanation:
NH3(aq) + H2O(l) ⇄ NH4⁺(aq) + OH⁻(aq)
I 0.95 0 0
C -x +x +x
E 0.95-x x x
Kb= [NH₄⁺] [OH⁻] / ( NH₃) = x²/ (0.95-x )
P(OH) = 14-PH = 14-11.612 = 2.388
(OH)⁻¹ = 10⁻²°³⁸⁸ = 4.09 x 10⁻³ = x
Kb = (4.09 x 10⁻³)²/ (0.95-4.09 x 10⁻³)
= 1.8 x 10⁻⁵
