3SnCl4 + 4Al(C2H5)3 3Sn(C2H5)4 + 4AlCl3?

Question

In one experiment, 0.170 L of SnCl4 (d = 2.226 g/mL) was treated with 0.347 L of triethylaluminum (Al(C2H5)3); d = 0.835 g/mL).
What is the theoretical yield in this experiment (mass of tetraethylstannane, Sn(C2H5)4)?
If 0.194 L of tetraethylstannane (d = 1.187 g/mL) were actually isolated in this experiment, what was the percent yield?

Answer 1

You have to get your reactants in terms of moles.
You have to determine which reactant is limiting
You then can find the percent yield.
1. 0.17 L*2226 g/L /mol wt SnCl2= moles of SnCl2 (call this "A")
2. 0.347 L*835 g/L/mol wt TEA=moles TEA
(call this "B").
From the equation, for each mole of SnCl4, you get 1 mole of product. and for each mole of TEA you get 3/4 mole product. So determine which compound, based on the actual moles creates the LEAST moles of TEA. Call this "C"
In the last part, compute 0.194L*1187g/L / mol wt TEStannate and call this "D".
The percent yield is 100*D/C