Upon combustion, a compound containing only carbon and hydrogen produces 2.49 g CO2 and 0.763 g H2O . Find the empirical formula
1 answer:
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Energy lost to condense = 803.4 kJ
<h3>Further explanation</h3>Condensation of steam through 2 stages:
1. phase change(steam to water)
2. cool down(100 to 0 C)
1. phase change(condensation)
Lv==latent heat of vaporization for water=2260 J/g
2. cool down
c=specific heat for water=4.18 J/g C
Total heat =
Answer:
From the degree of freedom analysis, the degree of freedom of the system and its components are equal to zero hence the system is well defined
Explanation:
To appraise the evaporation-crystallisation process, we go over the system to check if it is well defined from the available information as follows
45% by weight of inlet water is evaporated hence where inlet consists of 19.6% by weight of K₂SO₄ we have, Molar mass of K₂SO₄ = 174.259g/mol. Thus for every mole of K₂SO₄, we have 174.259×100/19.6 = 889.1g of solution is fed per mole of K₂SO₄,
Also the stream of concentrate leaving the evaporator contains
889.1 – 174.259 = 714.7 grams of water, and if 45% by weight of water is evaporated we have
45% of 889.1 is evaporated leaving a solution of weight = 889.1 × 55/100 = 489grams of solution which contains
100×174.259÷489 or 35.6% by weight of K₂SO₄ concentrate leaving the evaporator and moving on to the evaporator
However, 175. kg of water is evaporated/s hence from the previous calculation, quantity of water per mole of K₂SO₄ evaporated = 0.45×889.1= 400.1g which in comparison with actual quantity gives mass flow into
0.4001/175 = 0.003 or 437.39 to 1 hence the mass flow rate is 889.1g×437.39 = 388884g/s or 388.9Kg/s
a. Degrees of freedom analysis for the overall system
We have the following 4 unknowns in the overall system viz
m1, m3, m4 and m5
where m1 = maximum mass flow rate into the system
m3 = mass rate of evaporated water
m4 = maximum mass of solid K₂SO₄ crystals produced from the crystallizers
m5 = recycle ratio
While we have
1) Information, maximum capacity of evaporation from where we can calculate the maximum rate of feed supply
2) information, including chemical formula, to determine the maximum production rate
3) Information to calculate the water evaporated from fraction of water which is evaporated to that which is supplied
4) information to calculate the recycle ratio
Hence degrees of freedom = 4 – 4 =0
b. Degree of freedom analysis for the recycle-fresh feed mixing point
For the recycle-fresh feed mixing point we have m1 and m5, two unknowns
Where m1 is maximum feed rate and m5 is the mass of filtrate and we are given the compound molecular formula and the maximum flow rate from where we can calculate both m1 and m5
Hence the degrees of freedom = 0
c. Degree of freedom analysis for the evaporator
For the evaporator we have three unknowns m1,m2 and m3 and the available information are
1. The maximum water processing capacity of the evaporators and
2. The percentage quantity of water evaporated
Which is 2 hence we have 2 – 2 = 0 degrees of freedom
and
d. Degree of freedom analysis for the crystallizer the unknowns are m2, m4, m5
For the crystallizer the unknowns are m2, m4, m5The information available are
1. the ratio of crystals per kilogram of solution
2. The concentration of the recycled K₂SO₄ solution
3. Information of the maximum capacity of the evaporator so as to calculate the mass of concentrates leaving the evaporator and moving towards the crystallizer
Hence, we have 3 -3 = 0 degrees of freedom
From the degree of freedom analysis, the degree of freedom of the system and its components are zero hence the system is well defined
