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

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.

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.

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, 201115 yr by alpha2cen

You do realize that your reaction equation is wrong, right? Count the aluminum left and right.

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.

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)