chemistry dimensional analysis problem

Question

A piece of aluminum foil has a width of 12” and a thickness of 0.016 mm contains exactly 1.0 mole

of aluminum. How long is the piece? (The density of aluminum is 2.70 g/cm3).

Answer 1

D = m/v

A = lw

V = l x w x d(thickness)

l = V/w x d

12 in x 2.54 cm/ in = 30.48cm This is the width.

0.016 mm x 1cm/10mm = .0016 cm This is the depth

1.0 mole Al x 26.98g/1 mole Al = 26.98 g Al This is the mass.

D = m/v

2.70 g/cm^3 = 26.98 g/V

V = 26.98 g/2.70 g/cm^3

V = 9.99 cm^3

9.99 cm^3 = l x 30.48 cm x .0016cm

l = 9.99 cm^3/(30.48 cm)(.0016 cm)

l = 204.85 cm

You better check my math...