Question

For the following reaction, 9.42 grams of nitrogen monoxide are allowed to react with 8.55 grams of oxygen gas nitrogen monoxide (g) + oxygen (g) nitrogen dioxide (g) What is the maximum amount of nitrogen dioxide that can be formed? grams

Answer 1

1. 1. Nitrogen monoxide (g) + Oxygen (9) Nitra (NO) eitrogen dioxide (NO2). Balanced reaction is: 2NO + O2 – 200 Molembar weight of No = (14+16) = 30 g/mol Molecular weight of = (2x16) – 32 g/mol molecular weight of No₂=(19+32) = 46 g/mol In the reaction. 2 mol No reacts with I mol O2 to produce 2 mol NO2 , that is 60 g no reacts with 32g on to produce 929 NO2 50 32 g O2 reacts with big no to poroduce 1929 Noza . 8.155 g 0 reacts with 60 x 8.539 000 to produce a xests 9 No 9.42 g NO = 9.42 = 0.314 mol 8.559 O = 8.55 = 0.267 mol In the reaction 2 mol No peroduces 2 mol ie 92 g No2 92 x 0.314 g No, = 14.444 g NO2 Maximum 14.444 g NO₂ can be porodneed. 32 30 0.314 mol NO produces

2. 2Ag NO₂ (ar) + Nazco, (as) Ag, C03 (5) +2 Nano & excess Na CO3 66 mL 0.517 (M) AgNO3 1000 mL AgNO3 solution contain 0.517 mol Agnoz 66 mL AgNO3 solution centain 0.517... x66 mol Agro, 1000 = 0.034122 nol AgNO3 In the above reaction 2 mol AgNO3 produces 1 mol Ag, W3 0.034122 mol AgNO3 produces 1 40.034122 nol Ag2 403 0.034 122 x 275.74 g AgCO3 - 4.7 g Ag, CO3 molemlar weight of Ag CO₂ = (107.87 x2) +12 +48 - 275.74 g/rool 4.79 Ag coz will be precipitated.