Dom Posted January 21, 2011 Share Posted January 21, 2011 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 Link to comment Share on other sites More sharing options...
mississippichem Posted January 21, 2011 Share Posted January 21, 2011 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 Show how far you can get in the problem before you get stuck. I'll help pull you out of the rut. Link to comment Share on other sites More sharing options...
Dom Posted January 22, 2011 Author Share Posted January 22, 2011 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. Link to comment Share on other sites More sharing options...
alpha2cen Posted January 22, 2011 Share Posted January 22, 2011 (edited) 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. If conversion is 100%, first determine limit reactant. m_AlCl3/M_AlCl3=X mols m_NaOH/M_NaOH= Y mols X/Y<1/3 AlCl3 is limit reactant. if not X/Y < 1/3 NaOH is limit reactant. If AlCl3 is limit reactant, your solution is right.X*(78.1g/mol)is solution*reaction equation is wrong But if NaOH is limit reactant, Y/3*(78.1g/mol) is a solution. b) Depending on limit reactant. If AlCl3 is limit reactant remain NaOH = (Y-3X)M_NaOH If NaOH is limit reactant remain AlCl3 = (X-Y/3)M_AlCl3 Edited January 23, 2011 by alpha2cen Link to comment Share on other sites More sharing options...
Fuzzwood Posted January 22, 2011 Share Posted January 22, 2011 You do realize that your reaction equation is wrong, right? Count the aluminum left and right. Link to comment Share on other sites More sharing options...
mississippichem Posted January 22, 2011 Share Posted January 22, 2011 Dom, As Fuzzwood pointed out, your reaction equation is out of balance. Here is the balanced equation: [ce]AlCl_{3} + 3NaOH -> Al(OH)_{3} + 3NaCl[/ce] Now try to rework it and see what happens. Link to comment Share on other sites More sharing options...
Dom Posted January 22, 2011 Author Share Posted January 22, 2011 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) Link to comment Share on other sites More sharing options...
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