How many grams of Zn reacted in this reaction (very easy chemistry question)

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The discussion focuses on calculating the grams of zinc (Zn) that reacted in a reaction with hydrochloric acid to produce hydrogen gas. The initial calculations yielded 0.85 grams of Zn, but the textbook answer is 0.828 grams. The discrepancy arises from not accounting for the vapor pressure of water in the total pressure of the collected gas. By subtracting the vapor pressure of water (24.2 mm Hg at 25°C) from the total pressure, the correct partial pressure of hydrogen can be determined, leading to the textbook answer. Understanding the treatment of collected gas as either dry or wet is crucial for accurate calculations in such chemistry problems.
PhyIsOhSoHard
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Homework Statement


The reaction between zinc and hydrochloric acid is carried out as a source of hydrogen gas in the laboratory:
Zn(s)+2 HCl(aq)\rightarrow ZnCl_2(aq)+H_2(g)

If 325 mL of hydrogen gas is collected over water at 25°C at a total pressure of 748 mm Hg, how many grams of Zn reacted?

Homework Equations


PV=nRT
M=\frac{m}{n}


The Attempt at a Solution


Pressure in atmosphere:
\frac{748 mm Hg}{760 mm Hg/atm}=0.984 atm

Moles of Zn:
n=\frac{PV}{RT}=\frac{0.984 atm\cdot 0.325 L}{0.0821\frac{L\cdot atm}{mol\cdot K}\cdot (25+273)K}=0.013 mol

Grams of Zn:
m=M\cdot n=65.39g/mol\cdot 0.013 mol=0.85 g

I'm supposed to get 0.828 grams so I don't know what it is that I'm doing wrong?
 
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0.855 g (0.131 mol) is a correct answer.
 
Borek said:
0.855 g (0.131 mol) is a correct answer.

But my book's facit says 0.828 so I was worried maybe I had missed something.
 
PhyIsOhSoHard said:
I was worried maybe I had missed something.

Erratum :-p
 
Correct for vapor pressure of water

The problem mentions 'total pressure', so you have to take into account the vapor pressure of water. The vapor pressure of water depends on the temperature, and you can find that on a table.

Subtract it from the total pressure, to find the partial pressure of hydrogen. With this correction, your answer will match the textbook answer.
 
Good point, thank you.

Problem is, we never know at what level is the question asked, so it is rarely clear whether collected gas is to be treated as dry, or not (HS students know nothing about partial pressures). Apparently in this case 24.2 mmHg of water vapor does the trick.
 

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