A student used 0.1611 g of aluminum with excess HCl. This produced 205.1 mL of hydrogen gas. Atmospheric pressure was 754.7 torr and room temperature that day was 20.5 degrees Celsius?

Please help I am having a hard time. Can you please show all steps....
How many moles of hydrogen were produced?
What was the vapor pressure of the water at that temperature in torr?
What was the partial pressure of hydrogen in atm?
What value of R did the student obtain in L atm mol-1 K-1?

1 Answer
Jun 15, 2018

Answer:

Warning! Long Answer. Here's what I get.

Explanation:

It appears that you were collecting the hydrogen over water.

Doc Brown's Chemistry

Then, the gas you collected was a mixture of hydrogen and water vapour.

When you equalized the water levels inside the trough and the measuring cylinder, the total pressure of the two gases equalled the atmospheric pressure.

#p_text(atm) = p_text(H₂) + p_text(H₂O)#

There is no way that you could calculate the partial pressure of the water. Your instructor would either have given you the number or directed you to a table of vapour pressures like the one here.

Here's part of the table.

#ulbb(T"/°C"color(white)(m)p"/(torr)")#
#color(white)(m)20color(white)(mmm)17.5#
#color(white)(m)21color(white)(mmm)18.7#

You interpolate between the two pressures to get the pressure at 20.5 °C.

#"18.7 - 17.5 = 1.2"#
#0.5 × 1.2 = 0.6#
#17.5 + 0.6 = 18.1#

#p_text(H₂O) = "18.1 Torr"#

and

#p_text(H₂) = p_text(atm) - p_text(H₂O) = "(757.7 - 18.1) Torr =738.9 Torr"#

In atmospheres

#p_text(H₂) = 738.9 color(red)(cancel(color(black)("Torr"))) × "1 atm"/(760 color(red)(cancel(color(black)("Torr")))) = "0.9722 atm"#

We can use the Ideal Gas Law to calculate the moles of #"CO"_2#

#color(blue)(bar(ul(|color(white)(a/a)pV = nRTcolor(white)(a/a)|)))" "#

We can rearrange this formula to get

#n = (pV)/(RT)#

You need the value of #R#.

Your instructor would either have given you the value or told you to look it up your text.

In this problem,

#pcolor(white)(l) = "0.9722 atm"#
#V = "205.1 mL" = "0.2051 L"#
#R = "0.082 06 L·atm·K"^"-1""mol"^"-1"#
#T = "(20.5 + 273.15) K = 293.65 K"#

#n = ( 0.9722 color(red)(cancel(color(black)("atm"))) × 0.2051 color(red)(cancel(color(black)("L"))))/(0.082 06 color(red)(cancel(color(black)("L·atm·K"^"-1")))"mol"^"-1" × 293.65 color(red)(cancel(color(black)("K")))) = "0.008 274 mol" = 8.274 × 10^"-3"color(white)(l)"mol"#

You collected #8.274 × 10^"-3"color(white)(l)"mol"# of hydrogen.