Consider the subsequent balanced reaction. Explain how many grams of water are required to form 75.9g of HNO3?
Presume that there is excess NO2 present. The masses are as follows: 3 NO3 (g) + H2O yields 2HNO3 (aq) + NO (g)
H2O= 18.02g/mol HNO3= 63.02g/mol
Ok therefore I did not understand how to solve this. I tried 75.9g HNO3 x (1mol/63.02HNO3) x (2mol/1mol H2O) x (1mol/18.02gH2O)