Dom

Heres what I now have for my LR AlCl3 + 3Na(OH) - AL(OH)3 + 3 NaCL 15.0g 15.0g mAL(OH)3 1 mol AlCl3 reacts with 3 mol Na(OH) (133.33g/mol) (40g/mol) m3Na(OH) = (15.0g) 3(40.0g/mol) __________________ (133.33g/mol) = 13.5g In conclusion my LR is Na(OH)

ok well I bealive this is a theoretical yield question. This is what I have come up with as my balanced equation; AlCl3 + 3Na(OH) - 3Al (OH)3 + 3NaCl a) 1 mol of AlCl3 produces 3 mol Al(OH)3 (133.33 g/mol) 3(78.01 g/mol) 15.0g mAl(OH)3 mAl(OH)3= (15.0g) 3(78.1g/mol) __________________ (133.33g/mol) = 26.3g * this is what I get but on my exam revue sheet its 945g.

I have a question on my exam revue that I am having a hard time with. Its probably really easy, but I can't figure it out. MY QUESTION When a solution containing 15.og of aluminum chloride is mixed with a solution containing 15.0g of sodium hydroxide, a double displacement reaction occurs. a) Predict the mass of aluminum hydroxide produced. b) What mass of the excess reagent remains unreacted? thanks for you help