Problem 5-77

Volume of hydrogen needed to fill the balloon in liter with 20% lost:

V(H2) = 4800 m3 ×	1000 L	×	100	= 6.00×106 L
	m3		80

The corresponding number of mole of H₂ at STP is:

6.00×106 L ×	1 mol H2	= 2.71×105 mol H2
	22.41 L

Mass of iron split required:

2.71×105 mol H2 ×	1 mol Fe	×	55.85 g Fe	= 1.5×107 g Fe
	1 mol H2		mol Fe

Mass of H₂SO₄:
note: if H₂SO₄ is 98% purity, you need more to obtain a stoichiometric reaction (then: 100/98 = 1.02)

2.71×105 mol H2 ×	1.02 mol H2SO4	×	98.08 g H2SO4	= 2.7×107 g H2SO4
	1 mol H2		mol H2SO4