Pre lab: Volume of One Mole of a gas A student obtained the following data to de
ID: 1009117 • Letter: P
Question
Pre lab: Volume of One Mole of a gas A student obtained the following data to determine the mole volume of a gas. Measurement Value Mass of test tube and contents before heating. 19.440g Mass of test tube and contents after heating 19.148 g 214.1 mL 21.9 degrees C 700.0 torr Volume of water displaced. Temperature of water Atmospheric pressure 1. What is the mass of the O2 produced? 2. How many moles of O2 were produced? 3. What is the partial pressure of water (see the table in introduction)? 4. What is the partial pressure of O2? (see example 1) 5. What is the volume of oxygen corrected to STP (eq 5)? 6. What is the volume occupied by one mole of O2 gas at STP in this experiment (eq 6)Explanation / Answer
2KClO 2KCl + 3O
What is the mass of the O2 produced?
mass of the O = 19.440g - 19.148g
mass of the O = 0.292g --------answer
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How many moles of O2 were produced?
number of moles of O = m(O) / molar m(O)
number of moles of O = (0.292g) / (31.999g mol¹)
number of moles of O = 0.009125 mol --------answer
what is the partial pressure of water and O2
according to table
partial pressure of (HO) at 21.9 = 19.715 torr --------answer
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partial pressure of (O) = 700 torr - P(HO) @ 21.9
partial pressure of (O) = 700 torr - 19.715 torr
partial pressure of (O) = 680.285 torr--------answer
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What is the volume of oxygen corrected to STP
volume of O = Volume of water displaced 214.1ml
volume of O = 214.1m (1L/1000ml)
volume of O = 0.2141 L
equation --- PV = nRT
R = PV / ( nT)
PV / ( nT) = PV / ( nT)
( 680.285 torr * 0.2141L) / ( 0.009125 mol* 295.05 K) = (760 torr)V / ( 1 mol* 273.15 K)
volume of oxygen V = 19.44L --------answer
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