Answer:
See explanation.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to realize that the NaCl solution must react with an aluminum-containing substance, say the hydroxide or any other salt, so that the following equation will take place:
[tex]3NaCl(aq)+Al^{3+}\rightarrow 3Na^+(aq)+AlCl_3(s)[/tex]
In such a way, given the volume of the NaCl solution, it must be necessary to know its concentration, in order to get moles of this salt, further use the 3:1 mole ratio of NaCl to AlCl3 and the molar mass of the latter (133.34 g/mol) in order to solve an stoichiometric setup like the following:
[tex]m_{AlCl_3}=V_{NaCl}*M_{NaCl}*\frac{1molAlCl_3}{3molNaCl} *\frac{133.34gAlCl_3}{1molAlCl_3}[/tex]
Besides, you must make sure the volume is in liters.
Best regards!
