Question

The reaction 2 H_2 O_2(aq) right arrow 2 H_2 O(I) + O_2 (g) is first order in H_2 O_2 and has a rate constant of 0.00790 s^-1 at 20 degree C. A reaction vassel initially contains 159 mL of 26% H_2 O_2 by mass solution (the density of the solution is 1.17 g/mL). The gaseous oxygen is collected over water at 20 degree C as it forms. What volume of O_2 forms in 86 seconds at a barometric pressure of 746.5 mmHg? (the vapor pressure of water at this temperature is 17.5 mmHg). Determine the mass of Ca(OH)_2 in 365 mL of an aqueous solution that has a pH of 12.04.

Answer 1

159 ml when converted to mass we get mass = vol x density = 159 x 1.17 = 186.03 g

initial H2O2 mass = (26/100) x 186.03 = 48.3678 g

reaction is first order with respect to H2O2

hence t = ( 1/k) ln ( a/a-x)    where a= initial amt = 48.3678 g

86 = ( 1/0.0079) ln ( 48.3678 /a-x)

a-x = 33 g

H2O2 moles after 86 sec = mass/molar mass = 33/34 = 0.9706

H2O2 moles initially = 48.3678 /34 = 1.36376

H2O2 moles reacted = 1.36376-0.9706 = 0.39316

O2 moles foremed = 2 x H2O2 moles ( as per coeffiensts) = 2 x 0.39316 = 0.7863

P of O2 = 746.5-17.5 =729 mm Hg = 729/760 = 0.9592 atm ,

PV = nRT     where T = 20C = 20+273 = 293 K

0.9592 x V = 0.7863 x 0.08206 x 293

V = 19.7 L is vol of O2

2) pH = 12.04 ,pOH = 14-12.04 = 1.96 , [OH-]= 10^ -pOH= 10^-1.96 = 0.010965 M

[Ca(OH)2] = 1/2 [OH-] = 1/2 x 0.10965 = 0.0054824   ( since 1 Ca(OH)2 gives 2 OH-)

Ca(OH)2 moles = M x V = 0.0054824 x ( 350/1000) =0.00191884

Ca(OH)2 mass = moles x molar mass = 0.00191884 x 74.093 = 0.142 g