Xque
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Zn + acid gives how much hydrogen gas?
Ok, I need some help here.
"How many litres of hydrogen will you get when 6.5 grams of zinc is disolved in an excess of acid. Temp: 0°c, 1 Atm."
I assume that were dealing with an "ordinary" acid here, that is, an acid that does not have the ability to oxidize such as HNO3, which means that we
have the following reaction:
Zn (s) + 2 HA (aq) --> ZnA (aq) + H2
But except from that I'm lost.
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Nerro
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Calculate how many moles of Zinc you have (6.5g / 65.39g/mol =~ 0.1moles of Zinc. Which means you get 0,1 moles of H<sub>2</sub> . Use PV = nRT to calculate the volume of the hydrogen gas.
To use PV = nRT covert it first to V = (nRT)/P and since you're performing the reaction at 1 atm V = nRT/ 1atm will suffice for your purposes.
n = amount of moles of gas.
R = gas constant (0.08206 L atm / Mol K
T = absolute temperature in Kelvins (degrees C + 273,15)
V = volume in L (you want to calculate this)
P = pressure in atm
so (6.5g/65.39g/mole)(0,08206 L atm / mole K)(273.15 K) = 2.228L
But bare in mind that PV = nRT is the IDEAL GAS LAW and dihydrogen is not an ideal gas. When the volume is this small you can use it without seeing a
lot of difference from reality.) If their would be more gas, more exotic temperatures and higher pressure this method would show to be imperfect.
[Edited on Sat/Dec/2005 by Nerro]
[Edited on Sat/Dec/2005 by Nerro]
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Magpie
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IIRC one mole of gas at STP (standard temperature and pressure) has a volume of 22.4 liters. Since I believe OC and 1atm are standard conditions you
don't need any correction. At these moderate conditions I wouldn't worry about any non-ideality, assuming you are doing work where only rough
accuracy is required.
The single most important condition for a successful synthesis is good mixing - Nicodem
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Xque
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Of course. You almost never use that formula, so I didn't even consider it. Thank you!
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