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266,86 g aluminium chloride are obtained.
The answer is: approx. 327 g.
Silver chloride - AgClAg (107.89 grams) + Cl (35.45 grams) = 143.34 grams
Aluminum is 3+ and chloride is 1-. So, you need 3 chloride ions to neutralize 1 aluminum ion.
42,09 g silver chloride are obtained.
266,86 g aluminium chloride are obtained.
44.26% Aluminium Chloride is Aluminium.So there are 11.06g is aluminium in 25g.
g
The iron(III) hydroxide is not soluble in water and doesn't react with sodium chloride.
The empirical formula for aluminum chloride is AlCl3, and its gram formula mass is 133.34. The formula shows that each formula unit contains one aluminum atom, and the the gram atomic mass of aluminum is 26.9815. Therefore, 18(133.34/26.9815) or 89 grams, to the justified number of significant digits, of aluminum chloride will be produced.
The answer is: approx. 327 g.
4 moles
78 g
You need 145,337 g silver nitrate.
The ratio in aluminum chloride is 1:3 aluminum to chloride ( AlCl3 or Al2Cl6 )
17.9 grams
Silver chloride - AgClAg (107.89 grams) + Cl (35.45 grams) = 143.34 grams