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Section 11 (48pts total/&pts per question) partial credit given for work shown, AND work is required for credit, even if corr
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Answer #1

2.1) The moles of the gases present are calculated:

n F2 = P * V / R * T = 3 atm * 2.25 L / 0.082 * (120 + 273) K = 0.20 mol

n Cl2 = 5.75 * 4.5 / 0.082 * (120 + 273) = 0.80 mol

The average molar mass of the final gas is calculated:

MM ave = Xi * MMi = (0.2 * 38) + (0.8 * 71) = 64.4 g / mol

2.2) The reaction that occurs is:

2 KClO3 = 2 KCl + 3 O2

It is necessary that the vapor pressure of the water at that temperature is 0.06 atm

The moles of O2 are calculated:

n O2 = (0.95 - 0.06) * 0.25 / 0.082 * (35 + 273) = 0.0088 mol

The mass of KClO3 is calculated:

m KClO3 = 0.0088 mol O2 * (2 mol KClO3 / 3 mol O2) * (122.55 g KClO3 / 1 mol) 0.719 g

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