Dom
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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)
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(133.33g/mol)
= 26.3g * this is what I get but on my exam revue sheet its 945g.
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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
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Chemistry Revue Questions
in Homework Help
Posted
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)
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(133.33g/mol)
= 13.5g
In conclusion my LR is Na(OH)