what total volume of gas must be collected to obtain .586 g of He?

Question

Helium is collected over water at 25*C and 1.00 atm total pressure.

Answer 1

This is easy, however, the gas was not collected at STP, we can't use the easy method (where 1 mole of gas is 22.4 liters).

First, determine # of moles of He:

0.586g/4.00g/mol = 0.1465mol

Now, get the gas equation into proper form for solving this:

PV = nRT ----> V = nRT/P

get everything into proper units:

25 C = 298 K (other terms already OK)

Now, plug into the equation:

V = (0.08206latm/molK x 0.1465mol x 298K)/1.00atm =

V = 3.58 liters (note that all the units cancel except liters)

Answer 2

Moles Ag = 15.5 g/ 107.868 g/mol=0.14369 mol Ag moles O2 = 0.14369 x a million / 4 =0.03592 mol O2 Partial tension of water at 25 tiers Celsius is 24 mmHg. 757 mmHg - 24 mmHg = 733 mmHg / 760 mmhg = .96447 atm V = 0.03592 x 0.08206 x 298.15 ok/ 0.96447 atm = .911 L