Question
Info needed mass of Al- 0.0352 temperature- 19.0 degrees C--> 292 degrees K Barometric pressure- 762.5 mm Hg--> 1.0032 atm Volume of H2 gas- 46.92 I'm a little confused maybe someone can help; i convertedeverything, set up an equation for reaction between aluminum andhydrochloric acid i had to do and now im stuck on thesequestions: 1) what was the vapor pressure of the water in the gassample? 2) what was the pressure of the dry hydrogen? 3) determine the experimental value of the gasconstant,R 4) determine the volume of the dry H2 gas at STP conditions.(which i know standard temp and pressure) but not sure how to comeabout 5) determine the experimental value of volume of dry H2 gas atSTP per mole of H2 Info needed mass of Al- 0.0352 temperature- 19.0 degrees C--> 292 degrees K Barometric pressure- 762.5 mm Hg--> 1.0032 atm Volume of H2 gas- 46.92 I'm a little confused maybe someone can help; i convertedeverything, set up an equation for reaction between aluminum andhydrochloric acid i had to do and now im stuck on thesequestions: 1) what was the vapor pressure of the water in the gassample? 2) what was the pressure of the dry hydrogen? 3) determine the experimental value of the gasconstant,R 4) determine the volume of the dry H2 gas at STP conditions.(which i know standard temp and pressure) but not sure how to comeabout 5) determine the experimental value of volume of dry H2 gas atSTP per mole of H2
Explanation / Answer
I am assuming from what you said that the H2 gas was collectedover water and that the equation is this 2 Al(s) + 6 HCl (aq) -> 2 AlCl3+ 3H2(g) 1)The vapor pressure of water you can look up in a table basedont eh temperature of the system. 19.0 degrees C= .0217 at vapor pressure of water. You need this because the vapor pressure adds to yourfinal pressure and can mess up your results. 2)convert mass Al to mole H2. .0352g/ 26.962g/mol=.0013 mol .0013*3/2=.002 mol H2 P+.0217*.04692 L=.0821*.002*292 P dry hydrogen=1 atm 3) To find the experimental value fo R just use the values youfound in PV=nRT 4) use the same number fo moles just use the values fo 273 and1 atm 5) divide the volume you got in 4 by the number of moleH2 which is .002 mol.