0.43 moles of M form 18.23 grams of MCL
1 mole of M forms (18.23/0.43) grams of MCL
1 mole of M forms 42.4 grams of MCL
but 1 mole of CL weighs 35.4 grams
which means,
42.4 grams of MCL contain 35.4 grams of CL
18.23 grams of MCL contain (35.4/42.4)*18.23 grams of CL
=15.22 grams of CL in 18.23 grams of MCL
number of moles of CL in the sample = 15.22/18.23
= 0.835 of CL in the sample
A sample of 0.43 moles of a metal M reacts completely with excess chlorine to form 18.23 grams of MCl.
How many moles of Cl are in the sample of MCl that forms?
2 answers
I don't agree with Piom.
Yes, 18.23 g MCl contain 15.22g (I have 15.24 but we won't quibble about that). Then 15.22/35.44 = 0.429 moles Cl. But I think there is an easier way to do it.
2M + Cl2 ==> 2MCl
0.43 moles M means 0.215 moles Cl2 (just 1/2 M) to form 0.43 moles MCl.
0.215 moles Cl2 = 0.430 moles Cl.(If we used 15.24 from above in the first calculation, then 15.24/35.45 = 0.4299 or 0.43 which agrees with my shorter method.)
As an added bit of info, if we use 0.835 moles Cl x 35.45 = 29.60 g which is more Cl than you had to start with of 18.23 MCl so 0.835 can't possibly be correct.
Yes, 18.23 g MCl contain 15.22g (I have 15.24 but we won't quibble about that). Then 15.22/35.44 = 0.429 moles Cl. But I think there is an easier way to do it.
2M + Cl2 ==> 2MCl
0.43 moles M means 0.215 moles Cl2 (just 1/2 M) to form 0.43 moles MCl.
0.215 moles Cl2 = 0.430 moles Cl.(If we used 15.24 from above in the first calculation, then 15.24/35.45 = 0.4299 or 0.43 which agrees with my shorter method.)
As an added bit of info, if we use 0.835 moles Cl x 35.45 = 29.60 g which is more Cl than you had to start with of 18.23 MCl so 0.835 can't possibly be correct.
1 answer