Answer:
60 moles of NaF
Explanation:
The balanced equation for the reaction is given below:
Al(NO3)3 + 3NaF —> 3NaNO3 + AlF3
From the balanced equation above,
3 moles of NaF reacted to produce 1 mole of AlF3.
Therefore, Xmol of NaF will react to produce 20 moles of AlF3 i.e
Xmol of NaF = 3 x 20
Xmol of NaF = 60 moles
Therefore, 60 moles of NaF are required to produce 20 moles of AlF3.
density of a piece of metal = 7 gr/ml
See the file please
80.0 g Na and 20.0 g N2.
This means the limiting reactant determines the maximum mass of the product formed.
