For the following reaction, 6.73 grams of hydrochloric acid are mixed with excess barium hydroxide. The reaction yields 15.5 grams of barium chloride. 2 HCl (aq) + Ba(OH)2 (aq) = BaCl2 (aq) + 2 H2O (l) (1) What is the theoretical yield of barium chloride ? grams (2) What is the percent yield for this reaction ? %
step 1. 2 moles HCL -----> 1 mole BeCl2 step 2. find gram molecular wts of HCl and BeCl2
@toxicsugar22 now, express the answer
thsi is my first time doing this so please bear with me
im just practicing
no problem.. try it. it is fairly easy,...
just answer for step 2.
gram mol weight (gmw) = sum of atmic weights of all the elemnts in the molecule so, gmw of HCl = at. wt. of H + atomic wt. of CL
at. wt. of Cl = ? you have to refer to your Preiodic table
17 is the "atomic number" of Cl and 35.5 is the atomic wt.
1+35
good, so, gmw of HCl = 36.5 what this means is that 1mole of HCl weighs 36.5g (g=grams)
similarly, for BeCl2... notice that there are two atoms of chlorine
sorry it is BaCl2
practice to write it clearly.. gmw of BaCl2 = ___ + ___ + ____ = ____g
sorry I got 137 is the atomic number for Ba and 35 for Cl
no.. atomic number = number of protons in the nucleus of an atom atomic mass number = total number of protons + neutrons in the nucleus of an atom
ok so gmw of BaCl2 = ?
I added the two 137+ 35
good now back to step 2... summarize our values
@toxicsugar22
you see.. those are just numbers.. I cannot help you unless you put meaning to them. gram mol wt of HCL = gram mol wt of BaCl2 =
1mole of HCl weighs 36.5g 1 mole of BaCI2 WEIGHS 172
good.
now, using step 1, 2 moles of HCl ----------- yield ----------> 1mole BaCl2 i.e, 2* gmw of HCL ------------ yields -----------> 1*gmw of BaCl2
follow?
now, put your numbers
2*_ = __ g of HCl ---- > ____g BaCl2
perfecto..
02*36.5 = 73 g of HCl ---- > 172g BaCl2
so, 73g HCl ------> 172g BaCl2 (Theoretical) so, 6.73g HCl -------- should yield how much? (we have 6.73g HCl and Expected BaCl2 is?)
no.. HCl is our "reactant"..... BaCl2 is our product
no
ok, so, 73g HCl ----------- reacts to give us (yield) --------> 173g of BaCl2 how much HCl do we have in our reaction? (this part from the given question)
15.5 is the actual yield
just stay to the point... do not go jumping places.
6.73 is how much HCI WE HAVE
good.
so, if 73g HCl -------- provides -------> 173g BaCl2 then 6.73g HCl --------- should provide -----> ____g of BaCl2 this is a simple ratio..
follow?
STEPS!!!
..... \[{6.73 \times 173 \over 73}=?\]
15.94
ok. but do you understand this?
OK NOW i DO
I had to rethimk the problem
good. then THIS is your theoretical yield i.e., in theory, that 6.73g of HCl should give us with 15.94g of BaCl2
got it?
good thank you for tutoring me
not done yet that is just part (a)
part B. is where we do actual yield times theroicatl yield / 100
is that right
(b) % yield = (actual yield) * 100 / (theoretical yield) this value must be less than 100 actual yield = given in problem theoretical yield = calculated above
ok
so 15.5 times 100/ 15.94 =97.23
yes
ok good
can you go over one more problem with me
you are good tutor for me
just so I get this done evenfurthur
call it a day for today. some other time. keep practicing, put meanings to numbers and write clearly so you see what you are doing
ok thank you for helping me
@toxicsugar22 wtf not even a medal for the pain
soory I forgot
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