The depletion of ozone(O3) in the stratospere has been amatter of great concern
ID: 76597 • Letter: T
Question
The depletion of ozone(O3) in the stratospere has been amatter of great concern among scientists in recent years. Itis believed that ozone can react with nitric oxide(NO) that isdischarged from the high-altitude jet plane, the SST. The reactionis O3+NO--->O2+ NO2 If 0.740 g of O3 reacts with 0.670 g of NO, how many grams ofNO2 will be produced? Which compound is the limiting reagent?Calculate the number of moles of the excess reagentremaining at the end of the reaction. I need help with the whole problem but I would appreciateif I could get help on especially the bold part of theproblem----Calculate the number of moles of the excessreagent remaining at the end of the reaction. The depletion of ozone(O3) in the stratospere has been amatter of great concern among scientists in recent years. Itis believed that ozone can react with nitric oxide(NO) that isdischarged from the high-altitude jet plane, the SST. The reactionis O3+NO--->O2+ NO2 If 0.740 g of O3 reacts with 0.670 g of NO, how many grams ofNO2 will be produced? Which compound is the limiting reagent?Calculate the number of moles of the excess reagentremaining at the end of the reaction. I need help with the whole problem but I would appreciateif I could get help on especially the bold part of theproblem----Calculate the number of moles of the excessreagent remaining at the end of the reaction. I need help with the whole problem but I would appreciateif I could get help on especially the bold part of theproblem----Calculate the number of moles of the excessreagent remaining at the end of the reaction.Explanation / Answer
0.740 g O3 *(1mol O3/48 g) * (1mol NO/1mol O3) = 0.0154166667 molsNO required to react 0.670 g NO* (1mol NO/30g) = 0.0223333333 moles NO available So O3 is the limiting reactant 0.0223333333 moles NO - 0.0154166667 moles NO reacted =0.0069166666 moles NO left. EDIT: That should be round to 6.92 x 10-3 moles NO This has a mass of 0.0069166666 moles NO*(30 g/mol NO) = 0.207499998 grams˜ 0.207 grams NO (excess reactant) amount of NO2 formed 0.740 g O3 *(1mol O3/48 g) * (1mol NO2/1mol O3) *(46 g/mol) = 0.709grams NO2 formed