For 12.4g of liquid SoCl2 obtained from the reaction of 6.11g of gaseous So2 and excess gaseous Cl2, calculate the percent yield.

Question

step by step having trouble with percent yield do not seem to get the hang of it.

Answer 1

To find % yield, we need to know the actual yield and the theoretical yield. Then, the percent yield is the actual yield / theoretical yield (x100%).

We are given the actual yield, which is stated to be 12.4 g SOCl2. So, let us find the theoretical yield, i.e. the yield if you got 100% conversion to product.

Theoretical yield:

Reaction is SO₂ + 2Cl₂ ==> SOCl₂ + Cl₂O (assuming this is the reaction you refer to)

6.11 g SO₂ x 1 mol SO₂/64.1 g x 1 mol SOCl₂ / mol SO₂ x 119 g/mol SOCl₂ = 11.34 g SOCl₂ theoretical yield

percent yield = actual/theoretical (x100) = 12.4 g/11.34 g (x100) = 109% (impossible to have >100%)

(NOTE THAT THE ACTUAL YIELD IS GREATER THAN THEORETICAL. IMPOSSIBLE)

So, the only conclusion is that there is an error in the question of the reaction was meant to be

SO₂ + Cl₂ ==> SO₂Cl₂ and they meant to ask for the percent yield of SO2Cl2 instead of SOCl₂.