109/8.56=12.7
50+12.7
V=62.7
Mass= Volume x Density so i divided the mass and density to get the volume. and afterwards i would just add it to the mass to get my final answer
2 NH3+ 2 O2 —> 2 NO+ 3 H2O
S4O5
S3O
SeF6
N4S5
CCl9
All numbers should be subscripts
The Answer is D: Reactor products charge huge batteries.
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.
