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Nutka1998 [239]
3 years ago
8

What volume of so2 is produced at 325 k and 1.35 atm when 15.0 grams of hcl reacts with excess k2so3?

Chemistry
1 answer:
Helen [10]3 years ago
8 0
The balanced equation for the above reaction is as follows;
2HCl + K₂SO₃ ---> 2KCl + H₂O + SO₂
stoichiometry of HCl to SO₂ is 2:1
number of moles of HCl reacted - 15.0 g / 36.5 g/mol = 0.411 mol 
according to molar ratio 
number of SO₂ moles formed - 0.411 mol /2 = 0.206 mol
since we know the number of moles we can find volume using ideal gas law equation 
PV = nRT
where
 P - pressure - 1.35 atm x 101 325 Pa/atm = 136 789 Pa 
V - volume 
n - number of moles - 0.206 mol 
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature - 325 K
substituting values in the equation 

136 789 Pa x V = 0.206 mol x 8.314 Jmol⁻¹K⁻¹ x 325 K 
V = 4.07 L 
volume of SO₂ formed is 4.07 L


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An equilibrium mixture of PCl5(g), PCl3(g), and Cl2(g) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 Torr, respective
djverab [1.8K]

Answer:

The new partial pressures after equilibrium is reestablished for PCl_3:

P_1'=6.798 Torr

The new partial pressures after equilibrium is reestablished Cl_2:

P_2'=26.398 Torr

The new partial pressures after equilibrium is reestablished for PCl_5:

P_3'=223.402 Torr

Explanation:

PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)

At equilibrium before adding chlorine gas:

Partial pressure of the PCl_3=P_1=13.2 Torr

Partial pressure of the Cl_2=P_2=13.2 Torr

Partial pressure of the PCl_5=P_3=217.0 Torr

The expression of an equilibrium constant is given by :

K_p=\frac{P_1}{P_1\times P_2}

=\frac{217.0 Torr}{13.2 Torr\times 13.2 Torr}=1.245

At equilibrium after adding chlorine gas:

Partial pressure of the PCl_3=P_1'=13.2 Torr

Partial pressure of the Cl_2=P_2'=?

Partial pressure of the PCl_5=P_3'=217.0 Torr

Total pressure of the system = P = 263.0 Torr

P=P_1'+P_2'+P_3'

263.0Torr=13.2 Torr+P_2'+217.0 Torr

P_2'=32.8 Torr

PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)

At initail

(13.2) Torr     (32.8) Torr                        (13.2) Torr

At equilbriumm

(13.2-x) Torr     (32.8-x) Torr                        (217.0+x) Torr

K_p=\frac{P_3'}{P_1'\times P_2'}

1.245=\frac{(217.0+x)}{(13.2-x)(32.8-x)}

Solving for x;

x = 6.402 Torr

The new partial pressures after equilibrium is reestablished for PCl_3:

P_1'=(13.2-x) Torr=(13.2-6.402) Torr=6.798 Torr

The new partial pressures after equilibrium is reestablished Cl_2:

P_2'=(32.8-x) Torr=(32.8-6.402) Torr=26.398 Torr

The new partial pressures after equilibrium is reestablished for PCl_5:

P_3'=(217.0+x) Torr=(217+6.402) Torr=223.402 Torr

5 0
2 years ago
Please help me. Can you explain energy principle level?
Likurg_2 [28]
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8 0
3 years ago
How are a moon and an asteroid different? (2 points)
leonid [27]

Answer:

B

Explanation:

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8 0
3 years ago
The mass of 69Ga is 68.7200 amu . The mass of 71Ga is 70.9200 amu .
CaHeK987 [17]
I'm not sure if what this question is asking but ill be assuming that the average molecular mass is required. An assuming that the abundance of each isotope is 50% the average molecular mass is 69.82 amu. 
7 0
3 years ago
What is the solution to the problem to the correct number of significant figures (102,900/12)+(170•1.27)
GarryVolchara [31]

As per as the Multiplication rules of the significant figures, whenever any numbers in the decimals forms are multiplied or divided then result in mentioned in such a way so that the significant figures after the decimal will be same as that in the given least condition.


_______________________________


102900/12 = 8575


170 × 1.27 = 215.9


∴ (102,900 ÷ 12) + (170 × 1.27) =  8575 + 215.9


= 8790.9


Now, As per as Above rules, answer in correct significant figures will be = 8791.



8 0
3 years ago
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