phosphorus (12.28g) reacts with excess of oxygen gas to form 21.04 grams of diphosphorus penatoxide. what is the theoretical yile in grams of this reaction?
im not sure how exactly do you start this?
4 answers
*yield
You knew exactly what to do in the limiting reagent problem. These are simpler since there is only one reagent to worry about.
Convert 12.28 g phosphorus to moles. moles = grams/molar mass. This is the limiting reagent because you have only this reagent given and the other is "all you need."
Use the coefficients in the balanced equation to convert moles phosphorus to moles P2O5.
Then g P2O5 = moles P2O5 x molar mass P2O5. This is the theoretical yield in grams. Most problems that show a yield and ask for theoretical yield also want the percent yield. IN this case it is %yield = (actual yield/theoretical yield)*100 = ?
Convert 12.28 g phosphorus to moles. moles = grams/molar mass. This is the limiting reagent because you have only this reagent given and the other is "all you need."
Use the coefficients in the balanced equation to convert moles phosphorus to moles P2O5.
Then g P2O5 = moles P2O5 x molar mass P2O5. This is the theoretical yield in grams. Most problems that show a yield and ask for theoretical yield also want the percent yield. IN this case it is %yield = (actual yield/theoretical yield)*100 = ?
thanks, i got 74.77% :)
That looks ok to me.