[leyown]

 AlCl3 + 4NaOH
--> NaAlO2 + 3NaCl + 2H2O 
      How many grams of NaCl can be produced from 8.74 g of AlCl3 and 10.2 g of NaOH?

My question for this problem is how to calculate grams of NaCl since they gave us the mass for both reactants. 

[sjb]

AlCl3 + 4NaOH
--> NaAlO2 + 3NaCl + 2H2O 
      How many grams of NaCl can be produced from 8.74 g of AlCl3 and 10.2 g of NaOH?

My question for this problem is how to calculate grams of NaCl since they gave us the mass for both reactants. 

How many moles of aluminium chloride do you have? How many moles of sodium hydroxide?

[leyown]

The question gives you the mass of the two reactants and asks you to find the mass for NaCl.

What I did was I used 8.74g of AlCl3, converted it to moles, used a mole ratio between AlCl3 and NaCl, and then converted it back to grams. I got 11.5 g NaCl.

I did the same procedure except for the 10.2 g of NaOH and I got 11.2 g NaCl.

How do I know which one is the correct answer?

[sjb]

If you use up for instance 10 g of aluminium chloride in the reaction, how much sodium hydroxide would you need to react with it?

[leyown]

10 g AlCl3 x 1 mol / 133.18 g x 4 mol NaCl/1 mol AlCl3 x 58.4 g NaCl/1 mol NaCal

= 17.54 g NaCl

[AWK]

This is rather a very  bad example for the reaction with limiting reagent since this reaction proceeds only with an excess of NaoH.
Otherwise this reaction is important
AlCl₃ + 3NaOH = Al(OH)₃ + 3NaCl
And in this case all AlCl₃  forms NaCl

11.5 g is absolutely correct answer.

[DrCMS]

This is rather a very  bad example for the reaction with limiting reagent since this reaction proceeds only with an excess of NaoH.
Otherwise this reaction is important
AlCl₃ + 3NaOH = Al(OH)₃ + 3NaCl
And in this case all AlCl₃  forms NaCl

11.5 g is absolutely correct answer.

Yes this is the correct answer in the real world but is it the answer leyown's teacher will mark as correct?  Will they instead mark 11.2g as correct?