Chemistry Help!!!!?

Question

Hydrogen fluids is used in the manufacture of Freons (which destroys ozone in the stratosphere) and in the production of aluminium metal. It is prepared by reaction

CaF(2) + H(2) SO4 --> CaSO(4) + 2HF

In one process 6.00 kg of CAF(2) are treated with an excess of H(2) SO4 abd yield 2.86 kg of HF. Calculate the percent yield of HF.


Wow...i am so lost =[

Answer 1

Way above my comprehension. Maybe call one of your classmates up & ask them???

I would be lost too! LOL

Answer 2

I don't remember the ionization numbers, so you have to do that by yourself, but I can help. First off, convert 6 kg to 6000 grams.Then, make sure your original equation is balanced [CaF(2) + H(2) SO4 --> CaSO(4) + 2HF]. You have 1 calcium on one side along with 2 fluorines, and 2 hydrogens, one sulfur, and 4 oxygens. On the other side you have 1 calcium, one sulfur, 4 oxygens, 2 hydrogens, and 2 fluorines.This is balanced. So you take that 6000 grams of CAF(2) and divide it by CAF(2)'s molar mass (the mass of calcium times 2, plus the mass of fluorine times 2). You have that many moles of CAF(2). Get it? Then you multiply by the molar ratio (ratio of what you are trying to find over CAF(2). So basically, you multiply the moles of CAF(2) by 2. You see, because you have 2 moles of HF over 1 mole of CAF(2)). Then, you will have your moles of HF(call this answer number 1). Then, take that 2.86 kg that is given to you in the problem, and turn it into 2860 grams. You divide that by the molar mass of HF, which I know is around 20 (you add the masses of hydrogen and fluorine). Call this answer number 2. You then have your theoretical and actual yields of HF. Sound familiar? This is the equation for percent yield. You take answer number 2 and divide by answer number 1. Multiply by 100, and you have your answer.

Answer 3

hey!
the equation is 1 mole of calcium fluoride yields 2 moles of HF
so 1 mole of CaF(2) = 40 + (2 X 19) = 78 grams
i calculated these from the atomic weights

2 moles of HF= 2( 1+19)= 40 grams
now that data given is 6 kg of calcium fluoride is present = 6000 g

so if 78 g of CaF (2) yeilds 40 g of HF then 6000 g of CaF(2) will yeild x gm

x= 40 X 6000 / 78 = 3076 g = 3.076 kg

but actual yield is 2.86 kg

hence % of yield is (2.86 / 3.076) X 100 = 92.97%

mail me if u hav a doubt..i hope ive managed to make myself as clear as possible

Answer 4

alrighty, i did this last year so let's see.

you have your actual yield of 2HF so you need to calculate your theoretical yield.
i can't remember if the theoretical yield is started out with calculations by using the limiting reactant or the excess reactant

i feel like it's the limiting but i can't explain it to myself so i'm not sure.
if it is. you'll take the 6.00 kg and convert it to grams then divide my 78.08g of CaF(2) which is now converted to mols of CaF(2)

with those mols of CaF(2) you then multiply it by 2 because there are 2 mols of HF and only one of CaF(2) you then get your answer in mols of HF

take those mols and convert them first to grams then to kilograms
divide the 2.86 kg of actual yield by the number you just calculated which is the theoretical yield and multiply by 100 to get the percent yield. i got 93% when i did it. but like i said i really can't remember if it's the limiting reactant you use to calculate it.

i feel like it's the excess because that would account for the actual yield being smaller, but i'm being lazy. so if you think it is the excess i can calculate that, if i don't fall asleep.

Answer 5

Sorry i have no idea i passed chemistry with a c just last semester and i already forgot it all