20.0 mL of a 0.80 M aqueous lead(II) nitrate solution is mixed with an excess of an aqueous sodium sulfate solution. If 1.6 g of lead(II) sulfate is isolated from the reaction mixture, what is the percent yield of lead(II) sulfate?

%yield of is 33%

Explanation:

20.0 mL of 0.80 M = moles of

                                             = 0.016 moles of

                                             = 0.016 moles of

Balanced reaction:

According to balanced equation, 1 mol of produces 1 mol of

So, 0.016 mol of produces 0.016 mol of

As sodium sulphate remains in excess amount therefore lead nitrate is the limiting reagent.

So, theoretical yield of =  0.016 mol

Molar mass of = 303.26 g/mol

So, theoretical yield of =

%yield of = [(actual yield)/(theoretical yield)]%

                           =( ) %    

                           = 33%

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2

the answer is 0.42 is that is an option. this question provides no info