Chemistry HW Help: Calculating H2 Gas Yield from Zn and HCl at 825mm Hg & 42*C

[regnar]

124.7mL of a 6.0M solution of HCl is placed in a container with 25.8 grams of zinc metal. at 825mm Hg and 42*C, what volume of H2 gas can be produced from this reaction?


PV=nRT; R=0.0821 L*atm/mols*K



I don't if this is right but I decided to convert 25.8g of Zn into moles and then used the coefficients of the reaction and got 0.395mols Zn. Now, I don't know where to go from there if that is right.

 

[danago]

You are given the quantities of two reactants, so you will need to figure out which is the limiting reagent. Unless the amounts of HCl and Zn have been specifically calculated to be in stoichiometric proportions, you will end up with some excess of one of the reactants. You should then use the quantity of the limiting reagent for further calculations.

 

[regnar]

Oh, I'm sorry I forgot to put the reaction:

2 HCl + Zn ==> ZnCl + H2

 

[danago]

Double check your formula for zinc chloride, should be ZnCl₂.

With the equation, can you work out the limiting reagent? Calculate the number of moles of both Zn and HCl and then compare their ratio to the ratios indicated by the equation i.e. 2:1. Which one will be in excess?

 

[regnar]

I found HCl to be in excess and Zinc to be the limiting reagent.

 

[danago]

How did you come to that conclusion?

n(Zn) = 25.8 / 65.4 = 0.394 moles
n(HCl) = 0.1247*6 = 0.748 moles

According to the balanced equation, the number of moles of HCl required is double the number of moles of Zn. If all of the Zn did react, then it would require 2*0.394=0.789 moles of HCl, but there isn't that much present i.e. HCl is the limiting reagent.

 

[regnar]

Oh, I thought because that Zinc had the least amount of moles therefore making it the limiting reagent, but your reasoning makes perfect sense. Now that we have the limiting reagent, where do we go from there? I was told that have to use PV=nRT and work our way to liters of H2.

 

[danago]

Yes it is very important that you use the balanced equation to work out the LR, it will not always simply be the species present in the lowest quantities.

You are correct in suggesting the use of PV=nRT. Given that we now know that 0.748 moles of HCl will react, you can use the chemical equation to find the number of moles of H₂ produced and hence its volume under the specified conditions (using the ideal gas equation).

 

[regnar]

So, I find the moles of H2 and then plug that in for "n" in the equation and solve for V and that should be my answer?

 

[danago]

So, I find the moles of H2 and then plug that in for "n" in the equation and solve for V and that should be my answer?

Yep, pretty much. Be careful with units though; have a look at the units for the ideal gas constant, R, and then decide what units the pressure and volume should be in.

 

[regnar]

Yes, thank you. I converted everything to proper units before I did anything. Thank you for your help.