Current atmospheric pressure 21.34 .0y (Provided by instructure if in units othe

Question

Current atmospheric pressure 21.34 .0y (Provided by instructure if in units other than torr) Current atmospheric pressure torr 1 M HCI Sample 1 Sample 2 Mass of sample,g Initial buret reading, mlL Final buret reading, mL Total vol. H2 collected, mL Barometric pressure, torr Difference in water levels, mm Correction for difference in water levels, torr Pressure of wet hydrogen gas in buret, torr Temperature of water bath, Water vapor pressure at water bath 0391 903869 1.00 mL 1.7 ml 5.2 ml 34.57ml39 mL 433 m 25.5 | 23.5 temperature, torr Pressure of dry hydrogen, torr Moles H2 collected (which are equal to the moles of Mg reacted), mol Mass Mg in sample, g % Mg in sample, %

Explanation / Answer

1 in Hg is equal to 25.4 mm Hg so 29.34 in Hg is equal to

29.34 * 25.4 = 745.236 mm Hg or torr, divide this by 760 mmhg to get 0.98 atm

Temperature of water is 25.5 C, you need to find the vapor pressure of water at this temperature, im getting a value from internet of 24.5 torr

P total = P atm = Pwater + P Hydrogen collected, p is for presure

745.236 = 24.5 + P hydrogen

P hydrogen = 745.236 - 24.5 = 720.736 torrs, change it to atm = 0.94833 atm

Now apply ideal gas law equation

PV = n R T, p is pressure, v is volume, n is moles, R is gas constant, T is temperature

we can calculate the number of moles rearranging the equation

n = PV / RT, volume comes from the statement 34.57 ml or 0.03457 L , temperature for trial 1 in kelvin is 25.5 + 273.15 = 298.65 Kelvin

n = 0.94833 * 0.03457 / (0.082 * 298.65 ) = 0.001338 moles of H2

I see you have the reaction

Mg + 2HCl ==== MgCl2 + H2

we can see that 1 mole of hydrogen needs 1 mole of Mg, so we theoretically need 0.001338 moles of Mg to produce the moles of H2 here

molar mass of mg is 24.305 g/gmol

mass of mg = moles * molar mass = 0.001338 * 24.305 = 0.03252 grams of Mg in the sample

% of Mg is 0.03252 / 0.0391 = 0.8317 * 100 = 83.17 %

Repeat calculations for trial 2

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