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, 4.70 g of NH3 reacts with 5.83 g of O2.
Which is the limiting reactant? The answer is O2
How many grams of NO form? The answer is 4.37
How many grams of the excess reactant remains after the limiting reactant is completely consumed? THIS IS MY QUESTION. 4 was my original answer, but my teacher said it was wrong.
2006-09-04 17:25:09 · 1 answers · asked by cheezzznitz 5 in Science & Mathematics ➔ Chemistry
The molar mass of NO is 30.01 g/mol. If 4.37 g form, then
(4.37 g NO) / (30.01 g/mol NO) = 0.146 mol NO formed
Because NH3 and NO are in a 1:1 stoichiometric ratio, the amount of NO formed equals the amount of NH3 consumed (0.146 mol).
The molar mass of NH3 is 17.04 g/mol. If 0.146 mol are consumed, then
(0.146 mol NO) * (17.04 g/mol NO) = 2.48 g NH3 consumed
The process began with 4.70 g of NH3. Subtracting 2.48 g leaves 2.22 g NH3.
2006-09-04 18:10:23 · answer #1 · answered by prometheian 1 · 0⤊ 0⤋