How many grams of water are required to form 75.9g of HNO3?
Consider the following balanced reaction. How many grams of water are required to form 75.9g of HNO3?
Assume that there is ecess 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 so i didnt understand how to solve this. i tried 75.9g HNO3 x (1mol/63.02HNO3)x(2mol/1mol H2O)x(1mol/18.02gH2O) and it was wrong and Im not sure why. Pleaseexplain and give me the answer.
Homework Answers
Step2 Moles of H2O= (1/2) Moles of HNO3 = (1/2)(75.9/63.02)
Step3 Mass of H2O= (1/2)(75.9/63.02)(18.02)=10.85 g
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Consider the following balanced reaction. How many grams of water are required to form 75.9 g of HNO3? Assume that there is excess NO2 present. The molar masses are as follows: H2O = 18.02 g/mol, HNO3 = 63.02 g/mol. 3 NO2(g) + H2O(l) → 2 HNO3(aq) + NO(g) Answer 26.5 g H2O 10.9 g H2O 43.4 g H2O 21.7 g H2O 38.0 g H2O
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