Stoichem of Chemistry

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One of the steps in the commercial process for converting ammonia to nitric acid is the conversion of NH3 to NO.

4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)
In a certain experiment, 2.10 g of NH3 reacts with 3.44 g of O2.

(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?

Jul 24th, 2015

SOLUTION:

(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

** P/S: Please remember to rate my answer later on once the answer is correct. I would be highly appreciate it. Thank you. **
Jul 24th, 2015

** P/S: Please remember to rate my answer later on once the answer is correct. I would be highly appreciate it. Thank you. **

Jul 24th, 2015

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