Small amounts of hydrogen gas are often produced by reaction of zinc with HCl; Zn(s) + 2 HCl(aq) ® H2(g) + ZnC?
PLEAAAASE walk me through this! I have been struggling with this question all day! I already converted celsius to kelvin and subtracter water vapor from total pressure but don't know what to do from there
thank you soooo much if you can help!! i truly appreciate it
Also the number of moles was not given. I tried using PV=nrt but the answer kept coming out wrong..
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 447.1 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 1.255 g of Zn with an excess of HCl?
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You can calculate the amount of H2 produced based on the amount of Zn that was reacted with acid.
1.255 g Zn x (1 mole Zn / 65.38 g Zn) = 0.01920 moles Zn
The balanced equation tells us that 1 mole of Zn will produce 1 mole of H2.
0.01920 moles Zn x (1 mole H2 / 1 mole Zn) = 0.01920 moles H2O produced
P H2 = 447.1 mm Hg - 19.8 mm Hg = 427.3 mm Hg (x 1 atm / 760 mm Hg) = 0.5622 atm
T = 22.00 C + 273.15 = 295.15 K
PV = nRT
V H2 = nRT / P = (0.01920 moles)(0.08206 L atm / K mole)(295.15 K) / (0.5622 atm) = 0.8272 L
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