One of the steps in the commercial process for converting ammonia to nitric acid is the conversion of NH3 to NO.
(a) Which is the limiting reactant?
(b) How many grams of NO form?
(c) How many grams of the excess reactant remains after the limiting reactant is completely consumed?
(a) moles of NH3 = 2.10 g / 17.031 g/mol = 0.123 moles
moles of O2 = 3.44 g / 32 g/mol = 0.1075 moles
Since 0.123 moles / 4 > 0.1075 moles / 5, O2 is the limiting reactant.
(b) moles of NO = 0.1075 moles / 5 x 4 = 0.086 moles
mass of NO = 30.01 g/mol x 0.086 moles = 2.58 g
(c) moles of NH3 remaining = 0.123 moles - 0.1075 moles / 5 x 4 = 0.037 moles
mass of NH3 remaining =0.037 moles x 17.031 g/mol = 0.63 g
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