OK, I get a fair amount more wing needed. Let's check our math:

L = rho/2*V^2*Clmax*S

where S is wing area in ft^2, Clmax is max coefficient of lift for the 2D foil, V is airspeed in ft/s, and rho is air density, 0.002378 slug/ft^3 at STP. This works fine in a wind tunnel with the section going to the walls and the boundary layer sucked off of the walls near the foil.

Put the whole thing on an airplane and a couple more effects come in:

First is 3-D flow. Air runs from the high pressure side to the low pressure side, causing tip vortcies and resulting in the lift distribution trailing off towards zero at the tips. Regardess of wing shape, the spanwise lift distribution will look like the classic elliptical shape with the Cl at the root approximating the the lift from the wind tunnel model and at the very tip, lift is zero. So, we have to put PI/4 (as an reasonable approximation) into the equation to get the lift for a real wing;

Then, we have make lift to carry tail down load. A guess is that it will be 5 to 10% of the wing's lift.

So now the lift required of the wing is:

1.05*W = pi/4*rho/2*V^2*Clmax*S so

S = 1.05*W/(pi/4*rho/2*V^2*Clmax)

When I do this, it indicates 150 ft^2 of wing area needed to stall clean at 50 mph.

Billski