Small amounts of hydrogen gas are often produced by reaction of zinc with HCl; Zn(s) + 2 HCl(aq) ® H2(g) + ZnC?
Small amounts of hydrogen gas are often produced by reaction of zinc with HCl;
Zn(s) + 2 HCl(aq) ® H2(g) + ZnCl2(aq)
How many liters of H2 gas, collected over water at a pressure of 587.8 mm Hg and temperature of 22.00 °C (The vapor pressure of water at this temperature is 19.83 mm Hg.) will be produced by the reaction of 1.668 g of Zn with an excess of HCl?
Update:Please walk me through this?? I tried using PV=nrt but I don't think that's the way to get it because all my answers are wrong.
the first thing i did was subtract 19.83 from 587.8 and then convert to atm. then i changed degrees celsius to kelvin but then i don't know what else to do...
PLEASE HELP I AM DESPERATE and thanks a tons for your help :)
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You're on the right track; you do want to use PV=nrt. The only thing you didn't mention is the number of moles (n).
This is found by 1.668 g Zn * (1 mol Zn / 65.409 g Zn) * (1 mol H2(g) / 1 mol Zn) = .025501 mol H2 gass. The 1:1 molar ratio of Zn to H2 is shown by your given equation.
Now filling in PV = nrt:
(587.8 mm Hg - 19.83 mm Hg)(1 atm/760 mm Hg)*V = .025501 mol H2 * .0821 L atm/mol K * 295.15 K
V = .8267 L H2(g)
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