# of atoms per mol = Avogadro’s # (6.022 x 10^23)
Number of mols = mass of substance / molar mass
73 g / 40.08 g = 1.8 mols of Ca in 73 grams
1.8 mols x avagadro’s # = 1.1 x 10^24 atoms in 73 grams of Ca
Answer:
The third one makes the most sense
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.
Pan 4: theyre the smallest and most broken down :)
