Answer:
Ionization and dissociation
Answer:
3.3167 moles Of AlCl3
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
3Ca + 2AlCl3 —> 3CaCl2 + 2Al
From the balanced equation above,
2 moles of AlCl3 reacted to produce 2 moles of Al.
Finally, we shall obtained the number of moles of AlCl3 that reacted to produce 3.3167 moles of Al as follow:
From the balanced equation above,
2 moles of AlCl3 reacted to produce 2 moles of Al.
Therefore, 3.3167 moles Of AlCl3 will also react to produce 3.3167 moles of Al.
Thus, 3.3167 moles Of AlCl3 is needed for the reaction.
The answer lies in the stoichiometry of the reaction. If u look at the number BEFORE the reagent u will see the ratios of the reagents.
i think the answer is B. They have low reactivity.Hope this helped (: