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Chemistry 9 Online
OpenStudy (toxicsugar22):

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 ? %

OpenStudy (anonymous):

step 1. 2 moles HCL -----> 1 mole BeCl2 step 2. find gram molecular wts of HCl and BeCl2

OpenStudy (anonymous):

@toxicsugar22 now, express the answer

OpenStudy (toxicsugar22):

thsi is my first time doing this so please bear with me

OpenStudy (toxicsugar22):

im just practicing

OpenStudy (anonymous):

no problem.. try it. it is fairly easy,...

OpenStudy (anonymous):

just answer for step 2.

OpenStudy (anonymous):

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

OpenStudy (anonymous):

at. wt. of Cl = ? you have to refer to your Preiodic table

OpenStudy (anonymous):

17 is the "atomic number" of Cl and 35.5 is the atomic wt.

OpenStudy (toxicsugar22):

1+35

OpenStudy (anonymous):

good, so, gmw of HCl = 36.5 what this means is that 1mole of HCl weighs 36.5g (g=grams)

OpenStudy (anonymous):

similarly, for BeCl2... notice that there are two atoms of chlorine

OpenStudy (anonymous):

sorry it is BaCl2

OpenStudy (anonymous):

practice to write it clearly.. gmw of BaCl2 = ___ + ___ + ____ = ____g

OpenStudy (toxicsugar22):

sorry I got 137 is the atomic number for Ba and 35 for Cl

OpenStudy (anonymous):

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

OpenStudy (anonymous):

ok so gmw of BaCl2 = ?

OpenStudy (toxicsugar22):

I added the two 137+ 35

OpenStudy (anonymous):

good now back to step 2... summarize our values

OpenStudy (anonymous):

@toxicsugar22

OpenStudy (anonymous):

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 =

OpenStudy (toxicsugar22):

1mole of HCl weighs 36.5g 1 mole of BaCI2 WEIGHS 172

OpenStudy (anonymous):

good.

OpenStudy (anonymous):

now, using step 1, 2 moles of HCl ----------- yield ----------> 1mole BaCl2 i.e, 2* gmw of HCL ------------ yields -----------> 1*gmw of BaCl2

OpenStudy (anonymous):

follow?

OpenStudy (anonymous):

now, put your numbers

OpenStudy (anonymous):

2*_ = __ g of HCl ---- > ____g BaCl2

OpenStudy (anonymous):

perfecto..

OpenStudy (toxicsugar22):

02*36.5 = 73 g of HCl ---- > 172g BaCl2

OpenStudy (anonymous):

so, 73g HCl ------> 172g BaCl2 (Theoretical) so, 6.73g HCl -------- should yield how much? (we have 6.73g HCl and Expected BaCl2 is?)

OpenStudy (anonymous):

no.. HCl is our "reactant"..... BaCl2 is our product

OpenStudy (anonymous):

no

OpenStudy (anonymous):

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)

OpenStudy (toxicsugar22):

15.5 is the actual yield

OpenStudy (anonymous):

just stay to the point... do not go jumping places.

OpenStudy (toxicsugar22):

6.73 is how much HCI WE HAVE

OpenStudy (anonymous):

good.

OpenStudy (anonymous):

so, if 73g HCl -------- provides -------> 173g BaCl2 then 6.73g HCl --------- should provide -----> ____g of BaCl2 this is a simple ratio..

OpenStudy (anonymous):

follow?

OpenStudy (anonymous):

STEPS!!!

OpenStudy (anonymous):

..... \[{6.73 \times 173 \over 73}=?\]

OpenStudy (toxicsugar22):

15.94

OpenStudy (anonymous):

ok. but do you understand this?

OpenStudy (toxicsugar22):

OK NOW i DO

OpenStudy (toxicsugar22):

I had to rethimk the problem

OpenStudy (anonymous):

good. then THIS is your theoretical yield i.e., in theory, that 6.73g of HCl should give us with 15.94g of BaCl2

OpenStudy (anonymous):

got it?

OpenStudy (toxicsugar22):

good thank you for tutoring me

OpenStudy (anonymous):

not done yet that is just part (a)

OpenStudy (toxicsugar22):

part B. is where we do actual yield times theroicatl yield / 100

OpenStudy (toxicsugar22):

is that right

OpenStudy (anonymous):

(b) % yield = (actual yield) * 100 / (theoretical yield) this value must be less than 100 actual yield = given in problem theoretical yield = calculated above

OpenStudy (toxicsugar22):

ok

OpenStudy (toxicsugar22):

so 15.5 times 100/ 15.94 =97.23

OpenStudy (anonymous):

yes

OpenStudy (toxicsugar22):

ok good

OpenStudy (toxicsugar22):

can you go over one more problem with me

OpenStudy (toxicsugar22):

you are good tutor for me

OpenStudy (toxicsugar22):

just so I get this done evenfurthur

OpenStudy (anonymous):

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

OpenStudy (toxicsugar22):

ok thank you for helping me

OpenStudy (anonymous):

@toxicsugar22 wtf not even a medal for the pain

OpenStudy (toxicsugar22):

soory I forgot

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