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

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.