Please help with 2-4. My Chemistry teacher is not good at explaining how things

Question

Please help with 2-4. My Chemistry teacher is not good at explaining how things are put together and he never mentioned how to calculate using the 16.5cm. Any help would be appreciated.

Student Name: Stoichiometric Measurement of a Gas Product, Prelab questions: For questions 1-4, show all calculations with proper unit analysis and significant figures. Circle your final answer to each math question I. What theoretical yield (grams) of hydrogen gas can be obtained from 0.0334g Mg and 15.00mL of 6.0 M HCI? 2. D0031 thcb 334q Hy.0.0121 olm 0.02 imol H2 2. If the atmospheric pressure is 0993atm and the height of the water column at the end of the experiment is 16.5cm, what is the total pressure of the gas (mmHg) in the cudiometer? 3. If the total pressure of the gas in the eudiometer at 20.5°C is 580.4mmHg, what is the partial pressure of the hydrogen gas (mmHg)? 4. If the partial pressure of the hydrogen gas is 513.2mmHg, the volume of gas is 15.4mL, and the temperature of the water is 20.5°C, what was the actual yield of hydrogen gas (grams)? 5. Which reagent in this experiment is corrosive, and what safety precautions must be ob

Explanation / Answer

given pressure = 0.993 atm, height of water column= 16.5cm= 16.5/100m =0.165m

10.3 m of water   equals 1 atm ( considering water density as 1000 kg/m3 and Mercury as 13600 kg/m3 and converting 1atm= 760mm Hg in to water column= (760/1000)m** 136000 kg/m3/1000=10.3 m of water column)

=0.165 m correspond to 0.165/10.3 = 0.016

pressure = total pressure- height of water column expressed in atm =0.993-0.016=0.977 atm

=0,977*760 mm Hg =742.52 mm Hg

2. at 20.5 deg.c, from Antoine equation for water, log Psat (mm Hg)= 8.07131-1730.63/(t+233.426)

t in deg.c, log Psat(mm Hg) at 20.5 deg,c= 8.07131-1730.63/(20.5+233.426)=18.02 mm Hg

this is the saturation pressure of water, since the gas is saturated with water.

Partial pressure of gas= total pressure- saturation pressure of water= 513.2-18.02 mm Hg =495.18 mm Hg

2. given partial pressure, pressure exerted by the gas if it alone occupies the entire volume of gas mixture, P= 513.2 mm Hg= 513.2/760 atm =0.675 atm, V= 15.4ml= 15.4/1000 L=0.0154 L, T= 20.5 deg,c= 20.5+273= 293.5 deg.c

R =0.0821 L.atm/mole.K

from n= PV/RT =0.675* 0.0154/(0.0821*293.5)=0.000431 moles

moles =mass/molar mass, mass= moles* molar mas =0.000431*2 gm =0.000863 gm

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