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mote1985 [20]
3 years ago
15

A chemist has one solution that is 80% acid and another solution that is 30% acid. how much of the second (30%) solution is need

ed to make a 400 l solution that is 62% acid?
Chemistry
1 answer:
Aleksandr [31]3 years ago
8 0

messed up oops !!!!!!!!!


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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.
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