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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:

4NH3(g)+5O2(g)→4NO(g)+6H2O(g)

In a certain experiment, 1.55 g of NH3 reacts with 2.05 g of O2.

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The mass of nitrogen monoxide is 1.54 g.

What is the amount of NO produced?

In this case we have to make use of stoichiometry to obtain the mass of the NO that is produced in the reaction. The balanced reaction equation is;

[tex]4NH_{3} (g) + 5O_{2} (g) ---- > 4NO(g) + 6H_{2} O(g)[/tex]

Mass of ammonia= 1.55 g

Number of moles of ammonia =  1.55 g/17 g/mol = 0.088 moles

Mass of oxygen =  2.05 g

Number of moles of oxygen =  2.05 g/32 g/mol = 0.064 moles

If 4 moles of ammonia reacts with 5 moles of oxygen

0.088 moles of ammonia reacts with 0.088 moles *  5 moles/ 4 moles

= 0.11 moles

Hence oxygen is the limiting reactant.

If 5 moles of oxygen produces 4 moles of nitrogen monoxide

0.064 moles of oxygen produces 0.064 moles * 4 moles / 5 moles

= 0.0512 moles

Mass of nitrogen monoxide = 0.0512 moles * 30 g/mol = 1.54 g

Learn more about limiting reactant:https://brainly.com/question/14225536

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