omg I finally figured it out. Im literally gonna cry lmao
!! its 1.875g of NO and 1.6875g of H
2O However I am not sure if I am correct because the text book doesnt tell me so if someone can verify for me that'd be nice. I did it by multiplying the stoicheometric ratios by the moles of O
2 (because O2 was the limiting reactant) and converting that into grams. so to find grams of NO formed I did .078125mol O
2 × (4NO)/5molO
2) x 30g/1mol NO = 1.875 and did the same thing to find water. but now for C I have to find how many grams of excess reactant remain after limiting reactant is completely consumed and I got .9367g of NH3 because .078125 mol of O
2 need .0625 mol of NH
3 to fully react. So I subtracted .0625 from the available amount of moles which is .1176 and got .0551mol NH3 left. and converted it into grams by doing .0551 x 17 = .9367g the # 17 being the molar mass of NH
3. so is this all correct? what do you guys think? how am I supposed to answer (D)? how do I show that my calculations are consistent with the law of conservation of mass?