Question

Be sure to answer all parts. The hydrogen-oxygen fuel cell is described in the text. For the questions below, assume the following: Assume that the air is 20 % 0, by volume and that all the 02 is consumed in the cell. The other components of air do not affect the fuel-cell reactions. Assume ideal gas behavior. (a) What volume of H2(g), stored at 25.0°C at a pressure of 169 atm, would be needed to run an electric motor drawing a current of 8.50 A for 3.00 hr? L (b) What volume in liters) of air at 25.0°C and 1.10 atm will have to pass into the cell per minute to run the motor ? L/min

that is all the information i was given

Answer 1

By Faraday's law

W = $\frac{EQ}{F}$

Total charge (Q ) = it

Given, i = 8.50 A

t = 3 × 3600 sec = 10800 sec.

So, Q =( 8.50 × 91800) C.

F = 96500 C.

E is equivalent mass of Hydrogen = 1.00 g

W = mass of hydrogen reacted

Hence, mass of hydrogen gas reacted

W = (1.00× 91800/96500)

= 0.95 g.

Moles of Hydrogen = ( mass /molar mass)

= (0.95/2.00) = 0.475 .

Given, Pressure (P) = 169 atm.

T = 25+273= 298 K.

R = 0.082 L-atm/mol.K

Volume of hydrogen gas (V) is unknown

Now, using ideal gas equation.

PV = nRT

Or,V = (nRT/P)

Or, V = ( 0.475 × 0.082× 298)/169

= 0.068 L.

So, volume of H2 gas stored = 0.068 L.

H2 + 0.5 O2 = H2O

So, moles of oxygen needed = (moles of hydrogen /2) = (0.475/2) = 0.2375

Now, Volume of oxygen needed at 1.10 atm

= (nRT/P)

= (0.2375 × 0.082 × 298 )/1.10

= 5.28 L.

20% of air is O2

Then, total volume of air = (5.28 ×100)/20

= 26.4 L.

Now total time = 3.00 hour

So, total time = 3.00 × 60 = 180 min.

Now, Rate of air passed

= (26.4/180)

= 0.147 L/min.