One way to remove nitrogen oxide from smokestack emissions is to react it with ammonia. 4nh3+6no=5n2+6h2o calculate a) the mass of water produced from 0.839 mol of ammonia. b) the mass of no required to react with 3.402 mol of ammonia. c) the mass of ammonia required to produce 12.0 g of nitrogen gas. d) the mass of ammonia required to react with 115 g of no.

a) 22.7 g H₂O; b) 153.1 g NO; c) 5.84 g NH₃; d) 43.5 g NH₃

a) Mass of water

= 0.839 mol NH₃ × (6 mol water/4 mol NH₃) × (18.02 g H₂O/1 mol H₂O)

= 22.7 g H₂O

b) Mass of NO

= 3.402 mol NH₃ × (6 mol NO/4mol NH₃) × (30.01 g NO/1 mol NO) = 153.1 g NO

c) Mass of NH₃

= 12.0 g N₂ × (1 mol N₂/28.01 g N₂) × (4 mol NH₃/5 mol N₂)

× (17.03 g NH₃/1 mol NH₃) = 5.84 g NH₃

d) Mass of NH₃

= 115 g NO × (1 mol NO/30.01 g NO) × (4 mol NH₃/6 mol NO)

× (17.03 g NH₃/1 mol NH₃) = 43.5 g NH₃

d: the voltage of the secondary coil is less than that of the primary coil.

im not suree sorry if wrong

warm front forms when a moist, warm air mass slides up and over a cold air mass. as the warm air mass rises, it condenses into a broad area of clouds. a warm front brings gentle rain or light snow, followed by warmer, milder weather.

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