B
The answer should be B.....
Assuming ammonia is the product of this reaction:
N2+3H2--->2NH3
2 mole N2 * (3 mol H2)/(1 mol N2)= 6 mol H2
You would need 6 mol of hydrogen gas to completely react with 2 mol of nitrogen.
Answer:
Option D. 5.5
Explanation:
The equation is this:
2A + 6B ⇒ 3C
With the amounts that we were given, let's determine which is the <em>limting reactant</em>
2 A reacts with 6 B
4 A will react with ( 4 .6)/2 = 12B
I have 11 B, so the limiting is B
6 B react with 2 A
11 B will react with (11 .2 )/6 =3.66 A
I have 4 A, so A is the excess.
6 B produce 3 C
11 B will produce ( 11 .3)/6 = 5.5C
Answer:
New volume V2 = 92.7 Liter (Approx)
Explanation:
Given:
V1 = 106 l
T1 = 45 + 273.15 = 318.15 K
P1 = 740 mm
T2 = 20 + 273.15 = 293.15 K
P2 = 780 mm
Find:
New volume V2
Computation:
P1V1 / T1 = P2V2 / T2
(740)(106) / (318.15) = (780)(V2) / (293.15)
New volume V2 = 92.7 Liter (Approx)
