Asked by Christine
A balloon is filled with hydrogen gas. reaction is Fe(s) + H2SO4(aq) ----> FeSO4(aq) + H2(g).
The volume of the ballon was 4800 m cubed 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 degrees celsius, a pressure of 1.0 atm during filling, and a 100% yield.
The volume of the ballon was 4800 m cubed 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 degrees celsius, a pressure of 1.0 atm during filling, and a 100% yield.
Answers
Answered by
DrBob222
4800 m^3 = 4.8E6 L but you should confirm that.
PV = nRT. T is 273 K. Substitute from the problem and solve for n = number of mols. Then add 20% (0.20) to that to make up for the 20% loss.
Convert mols H2 to mols Fe. 1 mol H2 requires 1 mol Fe according to the equation. Then g Fe = mols Fe x atomic mass Fe.
Convert mols H2 to mols H2SO4. 1 mol H2 requires 1 mol 100% H2SO4 according to the equation. Then grams H2SO4 = mols H2SO4 x molar mass H2SO4 if H2SO4 is 100%. It isn't so grams 100%H2SO4/0.98 = grams 98% H2SO4. Post your work if you get stuck.
PV = nRT. T is 273 K. Substitute from the problem and solve for n = number of mols. Then add 20% (0.20) to that to make up for the 20% loss.
Convert mols H2 to mols Fe. 1 mol H2 requires 1 mol Fe according to the equation. Then g Fe = mols Fe x atomic mass Fe.
Convert mols H2 to mols H2SO4. 1 mol H2 requires 1 mol 100% H2SO4 according to the equation. Then grams H2SO4 = mols H2SO4 x molar mass H2SO4 if H2SO4 is 100%. It isn't so grams 100%H2SO4/0.98 = grams 98% H2SO4. Post your work if you get stuck.
Answered by
Victor
Thanks
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