Question

The combustion of ammonia in the presence of excess oxygenyields NO2(g) and H2O(g):4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(g)The combustion of 43.9 mg of ammonia produces how many grams ofNO2(g)?A. 178B. 0.119C. 43.9D. 0.954E. 2.58

Answer 1

Answer:

There will be produced 0.119 grams of NO2 (option B)

Explanation:

Step 1: Data given

Mass of ammonia = 43.9 mg

Oxygen is in excess

Molar mass of NH3 = 17.03 g/mol

Step 2: The balanced equation

4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(g)

Step 3: Calculate moles NH3

Moles NH3 = Mass NH3 / molar mass NH3

Moles NH3 = 0.0439 g/ 17.03 g/mol

Moles NH3 = 0.00258 moles

Step 4: Calculate moles of NO2

For 4 moles NH3 we need 7 moles of O2 to produce 4 moles NO2 and 6 moles of H2O

For 0.00258 moles of NH3 we'll have 0.00258 moles of NO2

Step 5: Calculate mass of NO2

Mass of NO2 = moles NO2 * molar mas NO2

Mass NO2= 0.00258 moles * 46 g/mol

Mass NO2 = 0.119 grams

There will be produced 0.119 grams of NO2 (option B)