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2 years ago

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

1 answer:

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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Suppose you are studying the K sp of K C l O 3 , which has a molar mass of 122.5 g/mol, at multiple temperatures. You dissolve 4

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Answer:

The K sp Value is $K_{sp}=7.40$

Explanation:

From the question we are told that

The of $KClO_3$ is = 122.5 g/ mol

The mass of $KClO_3$ dissolved is $m_s = 4.0g$

The volume of solution is $V_s = 12mL = 12*10^{-3}L$

The number of moles of $KClO_3$ is mathematically evaluated as

$No \ of \ moles \ = \frac{mass }{Molar \ mass}$

Substituting values

$= \frac{4}{122.5}$

$=0.0327\ moles$

Generally concentration is mathematically represented as

$concentration = \frac{No \ of \ moles}{volume }$

For $KClO_3$

$Z= \frac{0.0327}{12*10^{-3}}$

$=2.72 \ mol/L$

The dissociation reaction of $KClO_3$ is

$KClO_3 \ ----> K^{+}_{(aq)} + ClO_3^-_{(aq)}$

The solubility product constant is mathematically represented as

$K_{sp} = \frac{concentration of ionic product }{concentration of ionic reactant }$

Since there is no ionic reactant we have

$K_{sp} = [k^+] [ClO_3^-]$

$= Z^2$

$= 2.72^2$

$K_{sp}=7.40$

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SO3 has how many regions of high electron density and how many bonded electrons when the octet rule is strictly obeyed?1.5; 142.

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Answer:

3 regions of high electronic density

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