One way to remove the pollutant nitrogen monoxide, NO, from mobile diesel exhaust is by reacting it with ammonia, NH, as seen in the balanced reaction below. What mass, in grams, of NH, is required to remove 57.0 grams of NO from the air? 4 NH3 + 6 NO → 5 N, + 6H2O

In this case, in agreement to the given chemical reaction, it is possible for us to calculate the mass of NH3 required to remove 57.0 g NO via the stoichiometry based off the 4:6 mole ratio between them:

[tex]m_{NH_3}=57.0g NO*\frac{1molNO}{30.01gNO}*\frac{4molNH_3}{6molNO} *\frac{17.03gNH_3}{1molNH_3} \\\\m_{NH_3}=21.6gNH_3[/tex]

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One way to remove the pollutant nitrogen monoxide, NO, from mobile diesel exhaust is by reacting it with ammonia, NH, as seen in the balanced reaction below. What mass, in grams, of NH, is required to remove 57.0 grams of NO from the air? 4 NH3 + 6 NO → 5 N, + 6H2O​

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One way to remove the pollutant nitrogen monoxide, NO, from mobile diesel exhaust is by reacting it with ammonia, NH, as seen in the balanced reaction below. What mass, in grams, of NH, is required to remove 57.0 grams of NO from the air? 4 NH3 + 6 NO → 5 N, + 6H2O