How many milliliters of hydrogen will be produced by the reaction of 23.699 g Zn at 24 degrees C and 846 mm Hg?

1 Answer
Jul 5, 2017

You have left out a few details..........I gets approx. 8*L of dihydrogen gas..........

Explanation:

We are oxidizing zinc metal in (likely!) hydrochloric acid.....

Zn(s) + 2HCl(aq) rarr Zn^(2+) + 2Cl^(-) + H_2(g)uarr

And given this stoichiometry, ............................

"moles of metal "-=" moles of dihydrogen"

"Moles of zinc"=(23.699*g)/(65.39*g*mol^-1)=0.362*mol.

And so we should generate 0.362*mol "dihydrogen gas"

Now a pressure of 846*mm*Hg is specified. Clearly, the person who has set the question HAS NEVER used a mercury manometer. One atmosphere of pressure will support a column of mercury that is 760*mm high. And thus we must reduce the given pressure reading to units of "atmospheres":

"Pressure"=(846*mm*Hg)/(760*mm*Hg*atm^-1)=1.113*atm

And then we solve the Ideal Gas equation.......PV=nRT

V=(nRT)/P=(0.362*molxx0.0821*(L*atm)/(K*mol)xx297.15*K)/(1.113*atm)

V="approx." 8*L, "i.e. 8000"*mL.....

Just on the question of pressure measurement, which is why I find it objectionable that someone should quote a pressure of 846*mm, we can use a mercury column to measure pressure at about 1*atm (i.e. 760*mm*Hg) or at VERY low pressures, i.e. P=0.01*mm*Hg, 5*mm*Hg etc.. You put a mercury manometer under a pressure of greater than one atmosphere, and I bet you will get mercury all over the bench, and all over the lab. This is a major clean-up job, which contract cleaners won't touch.