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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below left is a cylinder containing water. an ball with a mass of 21g and a volume of 15 cm3 is lowered into the water. sketch t

anyanavicka [17]

You didn’t show the cylinder containing water, so I created one that you can use as a model (see image).

The water level was originally at 37 mL.

Then you added the ball, and it displaced its volume of water.

The new volume reading is 52 mL, so

Volume of ball = volume of displaced water = 52 mL – 37 mL = 15 mL.