Question

Using the following reaction (depicted using molecular models), large quantities of ammonia are burned in the presence of a platinum catalyst to give nitric oxide as the first step in the preparation of nitric acid. Suppose a vessel contains 7.50 g of NH3, how many grams of O2 are needed for a complete reaction?4 NH3 + 5 O2 --------> 4 NO + 6 H2O

Answer 1

Answer:

17.65 grams of O2 are needed for a complete reaction.

Explanation:

You know the reaction:

4 NH₃ + 5 O₂ --------> 4 NO + 6 H₂O

First you must know the mass that reacts by stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction). For that you must first know the reacting mass of each compound. You know the values ​​of the atomic mass of each element that form the compounds:

• N: 14 g/mol
• H: 1 g/mol
• O: 16 g/mol

So, the molar mass of the compounds in the reaction is:

• NH₃: 14 g/mol + 3*1 g/mol= 17 g/mol
• O₂: 2*16 g/mol= 32 g/mol
• NO: 14 g/mol + 16 g/mol= 30 g/mol
• H₂O: 2*1 g/mol + 16 g/mol= 18 g/mol

By stoichiometry, they react and occur in moles:

• NH₃: 4 moles
• O₂: 5 moles
• NO: 4 moles
• H₂O: 6 moles

Then in mass, by stoichiomatry they react and occur:

• NH₃: 4 moles*17 g/mol= 68 g
• O₂: 5 moles*32 g/mol= 160 g
• NO: 4 moles*30 g/mol= 120 g
• H₂O: 6 moles*18 g/mol= 108 g

Now to calculate the necessary mass of O₂ for a complete reaction, the rule of three is applied as follows: if by stoichiometry 68 g of NH₃ react with 160 g of O₂, 7.5 g of NH₃ with how many grams of O₂ will it react?

$mass of O_{2} =\frac{7.5 g of NH_{3} * 160 g of O_{2} }{68 g of NH_{3} }$

mass of O₂≅17.65 g

17.65 grams of O2 are needed for a complete reaction.