kabamawekesa

is it, 7.9-1.89=6.01 moles unreacted. So, 6.01x 63.55g/mol= 382g left?

Uhm, No sorry. I have no idea what to do after that

So, since HNO3= 7.5 moles and Cu=7.9 moles(500g/63.55), that means HNO3 is the limiting reactant right? Determined by the number of moles and since HNO3 has less moles, it is the limiting reagent. But isnt there another way to find the moles of Cu? Can I take the moles found in HNO3 and multiply it by 1molCu/4molHNO3? Giving me, 7.5mol*1molCu/4mol HNO3= 1.875 mol?

Well, I was not sure which one to use since certain people online said to use one certain reaction . The two reactions I got was, Cu+4HNO3->Cu(NO3)2+2NO2+2H2O or HNO3+Cu->Cu(NO3)2+H2 which one is the correct one to use? Hi, I tried a method using Cu+4HNO3->Cu(NO3)2+2NO2+2H2O The method I tried was, I used the concentration and volume of HNO3 to find the number of moles. ( 3.0x2.5)=7.5mol then used the moles I found to find the moles of Cu(NO3)2. 7.5x 1moleCu(NO3)2/4HNO3= 1.875 moles then took the moles I found with the Molar mass of CuNO3 to get the mass. (1.875x187.57= 351.7 grams). I think I did something wrong,perhaps in finding the moles for Cu.

Hi, Im in Grade 11 Chemistry and I have a question. That I do not understand. 500g of copper metal is reacted with 2.5L of 3.0 mol/L nitric acid solution. Calculate how much of the copper metal remains after the reaction is complete. I tried some ways but I still do not get the answer, I am looking for. The answer is 320g of copper left unreacted. Thank you, KB