In the reaction between aluminum and hydrochloric acid, how many moles of hydrogen gas would be liberated when 3.612E-2 grams of aluminum are treated with excess hydrochloric acid?

Question

Moles of hydrogen gas liberated = mol

The atmospheric pressure was measured to be 0.9666 atm and the room temperature was 25.00oC. At this temperature, the vapor pressure of water is 23.80 torr. What would be the volume of the number of moles of hydrogen just calculated under these conditions? Enter the answer in mL.

Volume of hydrogen gas = mL

The syringe used in this experiment is calibrated to 60 mL. However, there is a 'dead volume' at the needle end of the syringe that is uncalibrated. This volume equals 1.20 mL. Bearing in mind that the syringe is upside down, what would the reading on the syringe be that corresponded to the volume you have just calculated?

Reading on the syringe = mL

Suppose that you have completed this week's experiment and have determined that the molar mass of aluminum is 24.80 g mol-1. What is the percent error of your experimental result?

Percent error of the experimental result = %

Answer 1

2Al + 6HCl ==> 3H2 + 2AlCl3

3.612E-2g Al/molar mass Al = moles Al.
Convert moles Al to moles H2 by
?mols Al x (3 moles H2/2 moles Al) = ? mols Al x (3/2) = x moles H2.

PV = nRT
P = 0.9666 atm - v.p. H2O @ 25C.
v. p. H2O = 23.80mm/760 = xx
V = ?
R = 0.08206 L*atm/mol*K
T = 25.00 converedt to K.

I don't understand about the syringe.

%error = [(23.80-atomic mass Al)/atomic mass Al]*100 = ?