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# PLZ help me with these crazy ap chem problems

Registered User Posts: 2 New Member
edited September 2007
Can you solve these problems step by step? Thanks a bunch.

1). 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 m^3 and the loss of hydrogen gas during filling was estimated at 20.%. What mass of iron splints and 99% (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.

2). A 2.45 L sample of O2(g) was collected over a liquid at a total pressure of 785 torr and 25°C. When the O2(g) was dried (vapor of that liquid removed), the gas had a volume of 1.75 L at 25°C and 785 torr. Calculate the vapor pressure of this liquid at 25°C
Post edited by muchograndechica on

## Replies to: PLZ help me with these crazy ap chem problems

• Registered User Posts: 513 Member
1. If you lose 20% of H2 while filling the balloon, then you need 5 times the volume of the balloon to completely fill the balloon. Now that you have the volume of H2 needed, use PV=nRT to find the number of moles of H2 needed, and that's how many moles of Fe and H2SO4 you need (since it's a 1:1 ratio for all reactants/products) From there, it's easy to find out the mass of each reactant
• Registered User Posts: 2 New Member
so I do (1)(4800*5)=(.08206)(273)n?
with n, I do n/55.85 gr for iron flint?

• Registered User Posts: 1,068 Senior Member
For 2, I haven't done this in a while but it seems like you need to do partial pressures and/or mole fractions. Use 22.4 L=1 mol as a conversion.
• Registered User Posts: 513 Member
Not exactly. 1st you need to convert that volume to liters, otherwise that R value is useless (1 dm^3=1L),and (10dm=1m). This is because the units of R are (atm*L/[mol*K]) Then when you find n, note that it's units are in moles, so you multiply by 55.85 not divide, because the units are 55.85grams/mole. Same thing for H2SO4 except you divide by .99 in the end to account for the 99%
• Registered User Posts: 513 Member
Yanks, that 22.4L=1mol doesn't work in this case because you're not in STP (273K and 1atm). I'll get the method up soon
• Registered User Posts: 269 Junior Member
1) I'm going to do a lot of rounding...but...you'll get the point

4800L x 1mol/22.4L = 214.286 mol of H2

Since you lose 20%, do 214.286 x .20 = 42.8

Add 214.286 + 42.8 = 257 mol of H2

For Iron:

257 mol H2 x 1 mol Fe/1 mol H2 = 257 mole Fe

257 mol Fe x 55.85 g Fe/1 mole Fe = 14,300ish grams Fe

For H2S04:

257 mol H2 x 1 mol H2So4/1 mol H2 = 257 mole H2SO4

257 mol H2SO4 x 98 g H2SO4/1 mole H2SO4 = 25,000ish grams H2SO4

I might be off by a decimal place or something...but other than that, i'm pretty sure i'm right...unless i interpreted the question differently

2)

I'll do later maybe.
• Registered User Posts: 513 Member
Clove, 4800m^3 does not equal 4800L. Look at my previous post above.
They want the balloon to be filled but they lose 20% in the process, so you don't multiply by 0.20 because that wouldn't even be close to filling the balloon
Think about it like this. If you wanted to fill a 5L tank with water but you lose 20% of the water that comes in, you have 80% of the water that is left. So you have to divide by .80 to account for this

Actually thanks for bringing this up. I've erred as well. You actually divide by 0.80 not 0.20 (or multiply by 5 as I've suggested) so you actually multiply the volume by 1.25
• Registered User Posts: 513 Member
Oh wait, I just saw that you added the 20%, but that's still a little off because if you lose 20% of 257 you get 257-(51.4)=205.6 which is not equal to the 214.3 moles that you got earlier. The best way is to realize that 80% of the gas is kept and just divide by 0.8
And you forgot to divide the mass of the H2SO4 by 0.99 because it says 99% H2SO4 by mass.

So to do this problem
convert 4800m^3 to liters
divide by 22.4L/1mol to find number of moles of H2, multiply that by 1.25 to account for the 20% loss, set that number equal to the number of moles of Fe and H2SO4 you need, then convert from moles to grams for each reactant, remembering to divide by 0.99 for H2SO4 to account for the 99% mass
• Registered User Posts: 269 Junior Member
my bad about the liters. i just did some bad math by assuming that since 1 ml = 1 cm^3 1 l would = 1 m^3 ...haha

and i did make a lot of silly mistakes.

gator is correct.
This discussion has been closed.