0.528344 gallons in a 2 liter
Answer:
Volume of ammonia produced = 398.7 dm³
Explanation:
Given data:
Volume of N₂ = 200 dm³
Pressure and temperature = standard
Volume of ammonia produced = ?
Solution:
Chemical equation:
N₂ + 3H₂ → 2NH₃
Number of moles of N₂:
PV = nRT
1 atm× 200 L = n× 0.0821 atm.L/mol.K × 273 K
n = 200 atm.L /22.41 atm.L/mol
n = 8.9 mol
Now we will compare the moles of ammonia and nitrogen.
N₂ : NH₃
1 : 2
8.9 : 2/1×8.9 = 17.8 mol
Volume of ammonia:
1 mole of any gas occupy 22.4 dm³ volume
17.8 mol ×22.4 dm³/1 mol = 398.7 dm³
If you think of it endothermic is when there is energy needed for the reaction to occur and exothermic is when the reaction releases energy
Step one write the equation for dissociation of AgNO3 and NaCl
that is AgNO3-------> Ag+ + NO3-
NaCl--------> Na+ + Cl-
then find the number of moles of each compound
that is for AgNO3 = ( 1.4 x10^-3 ) x 25/1000= 3.5 x10^-5 moles
Nacl= (7.5 x10^-4)x 60/1000= 4.5 x10^-5 moles
from mole ratio the moles of Ag+= 3.5 x10^-5 moles and that of Cl-= 4.5 x10^-4 moles
then find the total volume of the mixture
that is 25ml + 60 Ml =85ml = 0.085 liters
The Ksp of Agcl = (Ag+) (cl-), let the concentration of Ag+ be represented by x and also the concentration be represented by x
ksp (1.8 x10^-10) is therefore= x^2
find the square root x=1.342 x10^-5
Ag+ in final mixture is = moles of Ag+/total volume - x
that is {(3.5 x10^-5)/0.085} - 1.342 x10^-5=3.98x10^-4
Cl- in the final mixture is =(4.5 x10^-5 /0.085) - 1.342 x10^-5= 5.16 x10^-4
