Explanation:
The net equation will be as follows.
So, we are required to find
for this reaction.
Therefore, steps involved for the above process are as follows.
Step 1: Convert K from solid state to gaseous state
,
= 89 kJ
Step 2: Ionization of gaseous K
,
= 418 KJ
Step 3: Dissociation of
gas into chlorine atom
.
,
= 122 KJ
Step 4: Iozination of chlorine atom.
,
= -349 KJ
Step 5: Add
ion and
ion formed above to get KCl
.
,
= -717 KJ
Now, using Born-Haber cycle, value of enthalpy of the formation is calculated as follows.
= 89 + 418 + 122 - 349 - 717
= - 437 KJ/mol
Thus, we can conclude that the heat of formation of KCl is - 437 KJ/mol.
Answer:
0.48g
Explanation:
Given parameters:
Mass of zinc = 15g
Unknown:
Mass of hydrogen gas = ?
Solution:
To solve this problem, we need to state the balanced reaction expression first;
Zn + 2HCl → ZnCl₂ + H₂
Find the number of moles of Zn;
Number of moles =
=
= 0.24mol
From the balanced reaction expression;
1 mole of Zn will produce 1 mole of H₂
0.24mole of Zn will produce 0.24mole of hydrogen gas
Mass of hydrogen gas = number of moles x molar mass
= 0.24 x (2 x 1)
= 0.48g
The mass of carbon in 1 liter of mixture = 1.108 g
<h3>What is the mass of carbon in 1 liter of the mixture?</h3>
The mass of carbon in 1 liter of the mixture is determined as follows:
First the moles of gas is determined using the ideal gas formula:
n = (1 * 1)/(0.08205L * 298)
n = 0.0409 mole of total gas
mass of gas is then determined using the formula:
mass = 1 * 1.375
mass = 1.375 g
Let x = mass of CH₄ and y = mass of C₄H₁₀
x + y = 1.375 g
nCH₄ + nC₄H₁₀ = ntotat
moles = mass/molar mass
x + y = 1.695 => y = 1.695 - x
(x/molar mass of CH₄) + [(1.375 - x)/ molar mass C₄H₁₀ = 0.0409
x/16 + (1.375 - x)/58 = 0.0409
x = 0.380 g CH₄
y = 1.375 - 0.380
y = 0.995 g of C₄H₁₀
mass of C in CH₄ = 12/16 * 0.380 = 0.285
mass of C in C₄H₁₀ = 48/58 * 0.995 = 0.823
Mass of carbon in 1 liter of mixture = 0.285 + 0.823
Mass of carbon in 1 liter of mixture = 1.108 g
In conclusion, the carbon is the major component in the mixture.
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