We are oxidizing zinc metal in (likely!) hydrochloric acid.....
Zn(s) + 2HCl(aq) rarr Zn^(2+) + 2Cl^(-) + H_2(g)uarrZn(s)+2HCl(aq)→Zn2++2Cl−+H2(g)↑⏐
And given this stoichiometry, ............................
"moles of metal "-=" moles of dihydrogen"moles of metal ≡ moles of dihydrogen
"Moles of zinc"=(23.699*g)/(65.39*g*mol^-1)=0.362*mol.Moles of zinc=23.699⋅g65.39⋅g⋅mol−1=0.362⋅mol.
And so we should generate 0.362*mol0.362⋅mol "dihydrogen gas"dihydrogen gas
Now a pressure of 846*mm*Hg846⋅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*mm760⋅mm high. And thus we must reduce the given pressure reading to units of "atmospheres"atmospheres:
"Pressure"=(846*mm*Hg)/(760*mm*Hg*atm^-1)=1.113*atmPressure=846⋅mm⋅Hg760⋅mm⋅Hg⋅atm−1=1.113⋅atm
And then we solve the Ideal Gas equation.......PV=nRTPV=nRT
V=(nRT)/P=(0.362*molxx0.0821*(L*atm)/(K*mol)xx297.15*K)/(1.113*atm)V=nRTP=0.362⋅mol×0.0821⋅L⋅atmK⋅mol×297.15⋅K1.113⋅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.
