At 25c, k =0.090 for the reaction: h2o(g) + cl2o(g) 2hocl(g)calculate the concentrations of all species at equilibrium for eachof the following cases: (a) 1.0g h2o and 2.0g cl2o are mixed in a1.0-l flask.(b) 1.0 mol pure hocl is placed in a 2.0-l flask.

a) [H2O] = 0.05093 M

[Cl2O] =  0.01843 M

[HClO]  = 0.00914 M

b) [H2O] = 0.2174M

[Cl2O] = 0.2174M

[HClO] = 0.0652 M

Explanation:

Step 1: Data given

Temperature = 25 °C

K = 0.090

Step 2: The balanced equation

H2O(g) + Cl2O(g) ⇆ 2HOCl(g)

(a) 1.0g H2O and 2.0g Cl2O are mixed in a1.0-L flask.

moles H20 = 1.00 / 18.02 g/mol = 0.0555 moles

Moles Cl2O = 2.00 grams / 86.91 g/mol = 0.0230 moles

Step 3: The initial concentration

[H2O] = 0.0555 M

[Cl2O]= 0.0230 M

[HOCl] = 0M

Step 4: Calculate concentration at equilibrium

For 1 mol H2O we need 1 mol Cl2O to produce 2 moles of HOCl

There will react X of H2O and X of Cl2O, there will be produce 2X of HOCl

The concentration at the equilibrium is:

[H2O] = (0.0555-X) M

[Cl2O]= (0.0230-X)M

[HClO] = 2X

Kc = 0.090

Kc = [HOCl]² / [H2O][Cl2O]

0.090 = 4X² / (0.0555-X)(0.0230-X)

K = 0.090 = 4X² / (0.072-X)(0.018-X)

X = 0.0045 7

[H2O] = 0.0555 - 0.00457 = 0.05093 M

[Cl2O] = 0.0230 - 0.00457 = 0.01843 M

[HClO] = 2*0.00457 = 0.00914 M

b) 1.0 mol pure HOCl is placed in a 2.0-L flask.

H2O(g) + Cl2O(g) ⇆ 2HOCl(g)

Step 1: The initial concentration

[H2O] = 0 M

[Cl2O]= 0M

[HOCl] = 1.00 mol / 2L = 0.500 M

Step 2: The concentration at the equilibrium

 For 1 mol H2O we need 1 mol Cl2O to produce 2 moles of HOCl

There will be produced X of H2O and X of Cl2O, while there will be consumed 2X of HOCl

[H2O] = XM

[Cl2O]= XM

[HOCl] =  (0.500 -2X) M

Kc = 0.090 = (0.50 - 2X)² / X²

0.090 = (0.0025 -2 + 4X²) / X²

X =0.2174

[H2O] = 0.2174M

[Cl2O] = 0.2174M

[HClO] = 0.500 - 2*0.2174 = 0.0652 M

b is the correct answer

mark me brainliest and hope this

2b2t

explanation:

2b2t