The answer is #0.13 grams#.
First, let's start with the balanced equation
#Zn(s) + 2HCl(aq) -> ZnCL_2(aq) + H_2(g)#
Now, the gas collected in t he #36.6mL# tube is hydrogen mixed with water vapor. In order to get the amount of hydrogen (#H_2#), we must get its partial pressure in the mixture by using Dalton's law.
So, the total pressure collected over the water is equal to the partial pressure of water and the partial pressure of #H_2#.
#P_(coll e c ted) = P_(H_2) + P_(water)#, which means
#P_(H_2) = P_(co l l e c t ed) - P_(water) = 890.0 - 17.5 = 872.5 mmHg#.
Using the ideal gas law, #PV = nRT#, we can determine the number of moles of #H_2# produced in the reaction
#n_(H_2) = (PV)/(RT) = ((872.5/760) * 36.6 * 10^(-3))/(0.082 * (20 +273.15)) = 0.002 mol es# (notice the conversions of pressure, volume and temperature to atm, L, and K, respectively);
From the balanced equation we can see that we have a #1:1# mole ratio between #Zn# and #H_2#, which means that the number of moles of #Zn# is equal to
#n_(Zn) = 0.002# moles
Therefore, the mass of #Zn# (knowing that its molar mass is #65.4 g/(mol)#), is equal to
#m_(Zn) = n_(Zn) * 65.4 g/(mol) = 0.004 mol es * 65.4 g/(mol) = 0.13 g#
Here's a helpful link to better visualize this reaction:
http://employees.oneonta.edu/viningwj/modules/CI_limiting_reactants_zinc_HCl.html
