J.R. S. answered 03/28/22
Ph.D. University Professor with 10+ years Tutoring Experience
4.0 kg H2O / day x 5 days = 20 kg H2O
heat needed to boil 20 kg of water starting @25º = q = mC∆T
q = (20 kg)(4.184 kJ/kgº)75º) = 6276 kJ needed
molar mass butane = 58.1 g mole
100 g butane / container x 1 mole / 58.1 g = 1.72 moles butane / container
1.72 mol butane / container x 125.7 kJ / mol = 216 kJ / container
6276 kJ x 1 container / 216 kJ = 29 containers needed = 30 containers minimum (due to rounding errors)