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?
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.
To find the number of moles of Cl in the sample of MCl, we need to determine the molar ratio between M and Cl in the compound MCl.
From the given information, we know that 0.43 moles of M react to form 18.23 grams of MCl.
To calculate the molar ratio, we need to determine the molar mass of MCl. The molar mass of MCl can be found by adding the atomic masses of M and Cl. Let's assume that M has an atomic mass of x.
Since the compound MCl contains 1 mole of M and 1 mole of Cl, its molar mass will be the sum of the atomic masses of M and Cl:
Molar mass of MCl = x + 35.45 (the atomic mass of Cl)
Given that 18.23 grams of MCl are formed, we can use the molar mass equation to find the molar mass of MCl:
Molar mass of MCl = mass / moles = 18.23 g / 0.43 mol
Solving for the molar mass of MCl:
Molar mass of MCl ≈ 42.4 g/mol
Since the molar ratio between M and Cl in MCl is 1:1, if there are 0.43 moles of M, there will also be 0.43 moles of Cl.
Therefore, the sample of MCl that forms contains 0.43 moles of Cl.
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