Quiz 7 J Complete the following assignment on this sheet or separate paper (not
ID: 545768 • Letter: Q
Question
Quiz 7 J Complete the following assignment on this sheet or separate paper (not spiral wound paper). Your solutions should show all work presented neatly and in pencil. Make sure you clearly indicate your answers. Problem 1. The reaction 2 H2O2(aq) 2 H2O(l) + O2(g) is first order in H2O2 and has a rate constant of 0.00775 s-1 at 28 °C. A reaction vessel initially contains 254 mL of 29 % H2O2 by mass solution (the density of the solution is 1.06 g/mL). The gaseous oxygen is collected over water at 28 °C as it forms. What volume of O2 forms in 203 seconds at a barometric pressure of 763.4 mmHg? (the vapor pressure of water can be found in Appendix E) Problem 2. Determine the mass of Ca(OH)2 in 995 mL of an aqueous solution that has a pH of 10.48. 106 ovdo l vespl to 4201 0776Explanation / Answer
1)
2 H2O2 (aq) ------> 2 H2O + O2
mass of H2O2 = 254 * 1.06 = 269.24 g
mass of H2O2 = 269.24 * 29/100 = 78.096 g
moles of H2O2 = 78.096/34 = 2.297 moles
K = 2.303/t * log(a/(a-x))
0.0077 = 2.303/203 * log(2.297/x)
x = final number of moles of H2O2 = 0.4813 moles
reacted moles of H2O2 = 2.297 - 0.4813 = 1.8157 moles
2 moles of H2O2 produces 1 mole of O2
1.8157 moles of H2O2 produces 1/2 * 1.8157 moles of O2
number of moles of O2 = 0.9078 moles
pressure of O2 gas = 763.4 - aqueous tension
pressure of O2 = 763.4 - 28.3 mmHg = 735.1 mmHg
Pressure of O2 = 735.1/760 = 0.9672 atm
according to ideal gas equation PV = nRT
0.9672 * V = 0.9078 * 0.0821 * (28+273)
V = volume of O2 = 23.194 L
2)
pH = 10.48
pOH = 14 - pH
pOH = 14 - 10.48
pOH = 3.52
[OH-] = 10^-3.52
[OH-] = 3.02 * 10^-4 mol/L
M1V1 = M2V2
3.02*10^-4 * 1 L = M2 * 995/1000
M2 = 3.0352 * 10^-4 mol/L
mass of Ca(OH)2 = 3.0352 * 10^-4 * 74.09
mass of Ca(OH)2 = 0.02249 g
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