In 1897 the Swedish explorer Andree tried to reach the North Pole in a balloon. The balloon was filled with hydrogen gas. The hydrogen gas was prepared from iron splints and diluted sulfuric acid. The reaction is given below. Fe (s) + H2SO4 (aq) - ->FeSO4 (aq) + H2 (g). The volume of the balloon was 4800 m3 and the loss of hydrogen gas during filling was estimated at 20.%. What mass of iron splints and 98% (by mass) H2SO4 were needed to ensure the complete filling of the balloon? Assume a temperature of 0°C, a pressure of 1.0 atm during filling, and 100% yield.

Question

Chemistry

Duckling

3 January, 23:39

In 1897 the Swedish explorer Andree tried to reach the North Pole in a balloon. The balloon was filled with hydrogen gas. The hydrogen gas was prepared from iron splints and diluted sulfuric acid. The reaction is given below. Fe (s) + H2SO4 (aq) - ->FeSO4 (aq) + H2 (g). The volume of the balloon was 4800 m3 and the loss of hydrogen gas during filling was estimated at 20.%. What mass of iron splints and 98% (by mass) H2SO4 were needed to ensure the complete filling of the balloon? Assume a temperature of 0°C, a pressure of 1.0 atm during filling, and 100% yield.

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Answers (1)

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Kennedi Singh · Answer 1

Iron splints = 14352.50 Kg

98% H2SO4 = 25186.56 Kg

Explanation:

First, you need to know how much H2 gas you need to fill the balloon. If we consider H2 as an ideal gas, we can solve the problem using the ideal gas law written as pV=nRT (eq 1)

p = pressure (atm) = 1.0 atm

V = volumen=4800m3=4.3E6

n = number of moles of H2

R = ideal gas constant = 0.08205 (atm*L) / (mol*K)

T = Temperature = 0°C = 273.15 K

Then we solve the equation for moles of H2 and use the info to solve the eq (2)

n=PV/RT eq (2)

n = ((1atm*4.8E6) / (0.08205 (atm*L) / (mol*K))

n=214171.386 mol H2

If we considerer hydrogen loss of 20%, we need to produce 20% more of H2, so the real necessary ammount of hydrogen is n = (214171.386*1.2) = 257005.67 moles H2

According to chemical equation, reaction between 1 mole of Fe splints and 1 mole H2SO4 produce 1 mole of H2, so 257005.67 moles of H2 are produce by the same ammount of Fe and H2SO4 moles.

Next, we need to make the conversión of moles to mass using eq (3) and solving for mass

n = g/M. M eq (3)

Where n = moles, g = grams and M. M. is the molar mass of the compount (Fe=55.845 g/mol and H2SO4=98g/mol) so then:

g = n*M. M.

g of Fe = (257005.67 moles) * (55.845 g/mol) = 14352481.64 grams = 14352.50 Kg of Fe splints

g of H2SO4 = ((257005.67 moles) * (98 g/mol) = 25186555.66 grams = 25186.56 Kg of H2SO4

Finally, we should to considerer that if H2SO4 were 100%, we need 25186.56 kg, but the hydrogen gas was prepared using diluted acid (98%) which means that for every 100 Kg of diluted acid, 98 Kg are H2SO4 and 2 Kg are wáter, so: H2SO4 98% = 25186.56 * (100/98) = 25700.57 Kg