Question

Please complete these calculations and show all of the work required for understanding purposes.

DATA: Trial 2 Trial 1 Mass of aluminum sample Volume of Ha collected 0.0309 39,0ML 22°c 16/mmHg Temperature at gas-collection site Barometric pressure CALCULATIONS: 1. Vapor pressure of water (from table on page 51) 2. Pressure of "dry" H2 collected 3. Amount of Hz produced (mol) 4. Amount of Al reacted (mol) 5. Mass of Al in soda can sample 6. Percent by mass of Al in sample 7. Average percent of Al in sample

The sample is the mass of the aluminum

Answer 1

Pressure of dry gas = barometric pressure - vapor pressure of water

= (760 -19.8) mm Hg

= 740.2 mm Hg.

= (740.2/760) = 0.973 atm

Volume of gas = 39.0 mL = ( 39.0/1000) = 0.039 L.

Temperature of gas = 22⁰ c = 273+22= 295 K.

Using ideal gas equation

PV = nRT

Or, n = (PV/RT) = (0.973×0.039)/(0.082×295)

= 0.00157

Moles of hydrogen = 0.00157

Now,

Reaction for formation of hydrogen.

2 Al + 6 HCl $\to$ 2 AlCl₃ + 3H₂

so, moles of Al = $\frac{2}{3}$ × Hydrogen = ($\frac{2}{3}$ × 0.00157)

= 0.00104

Mass of Al = moles × molar mass

= 0.00104×27 = 0.0281 g.

Percentage of Al in the sample

= $\frac{0.0281 \times 100}{0.03}$

= 93.66 %.