1. Leave everything alone and don't touch the glass .
2. Tell a Teacher .
3. And that's it .
Answer- 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.
Given - Number of moles of Al(NO3)3 - 4 moles
Number of moles of NaCl - 9 moles
Find - Maximum amount of AlCl3 produced during the reaction.
Solution - The complete reaction is - Al(NO3)3 + 3NaCl --> 3NaNO3 + AlCl3
To find the maximum amount of AlCl3 produced during the reaction, we need to find the limiting reagent.
Mole ratio Al(NO3)3 - 4/1 - 4
Mole ratio NaCl - 9/3 - 3
Thus, NaCl is the limiting reagent in the reaction.
Now, 3 moles of NaCl produces 1 mole of AlCl3
9 moles of NaCl will produce - 1/3*9 - 3 moles.
Weight of AlCl3 - 3*133.34 - 400 grams
Thus, 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.
is this a multiple choice question?
<span>389.88094 amu
First we look up the atomic mass of all elements contained in calcium iodate using the periodic table:
Ca: 40.078
I: 126.90447
O: 15.999
As an intermediate step we calculate the molecular mass of the ion IO3:
126.90447 + 3*15.999 = 174.90147
Then we calculate the mass of one calcium atom and 2 iodate ions:
2*174.90147 + 40.078 = 389.88094 amu</span>
<span>When a sulfur atom gains its valence electrons to have 8, the charge on the resulting ion is stable since sulfur atom achieved the octet rule. An octet rule is the filling of 8 electrons in an atom.</span>
