When lead nitrate reacts whith sodium iodide, sodium nitrate and lead iodide are formed if i start with 25.0 grams of lead nitrate and 15.0 grams of sodium iodide, find the percent yield if this reaction forms 6.0 g of sodium nitrate during the experiment
pb(no3)2+ 2nal--> pbl2+ 2nano3

So you find the moles of the lead nitrate by doing mass/Mr so 25/331.2 which gives you 0.075

Due to stoichiometry in the reaction we see 1:2 ratio so we do 0.075x2 gives 0.15 moles of NaNO3

Then mass: moles x Mr so 0.15 x 85= 12.75g is your theoretical

So % yield is actual/ theoretical x 100

Therefore 6.0g/12.75 g x 100 = 47%

Not sure if it’s correct hope it helps!