Answer: 48.95g
Explanation:
no. of moles of Cl2 = 39/(2*35.5) = 0.55 mol
no. of moles of Al = 34/27 = 1.26 mol
hence, aluminium is in excess so we'll do calculation using no. of moles of Cl2 as it will be the only reactant to be used up completely. So,
no of moles of AlCl3 = 2/3 * (0.55) = 0.367 mol
hence amount of AlCl3 = 0.367 * (27+3*35.5) = 48.95g
Answer:
318 g / 19.32 g v = 16. your volume is 16 hope this helps
Explanation:
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