Question

Nitric acid, a major industrial and laboratory acid, is produced commercially by the multi-step Ostwald process, which begins with the oxidation of ammonia: Step 1: NH3(g) + O2(g) LaTeX: \longrightarrow⟶ NO(g) + H2O(l) Step 2: NO(g) + O2(g) LaTeX: \longrightarrow⟶ NO2(g) Step 3: NO2(g) + H2O(l) LaTeX: \longrightarrow⟶ HNO3(l) + NO(g) Assuming 100% yield in each step, what mass (in kg) of ammonia must be used to produce 7.839 kg of nitric acid. Enter to 3 decimal places. Are the equations balanced?

Answer 1

• Answer:  3.178 Kg of ammonia must be used to produce 7.839 kg of nitric acid and The equations are not balanced.  Explanation:  Ostwald process  is a multi-step for manufacturing Nitric Acid.  First, We must check if our equations are balanced or not, by checking the amount of each element before and after the arrow.  Step 1: NH3(g) + O2(g) ⟶ NO(g) + H2O(l)  Step 2: NO(g) + O2(g) ⟶ NO2(g)  Step 3: NO2(g) + H2O(l) ⟶ HNO3(l) + NO(g)  For example in Step 1, the amount of Hydrogen atoms are different on both sides. On left side we can count 3 hydrogen atoms while on the right side we can count only 2 hydrogen atoms. In the Step 2, the amount of Oxygen atoms are different on both sides too. On the Left side we can count 3 oxygen atoms while on the right side we can count only 2 oxygen atoms and in step 3 the amount of Nitrogen atoms are different on both sides too. On the left side we can count 1 nitrogen atom while on the right side we can count 2 nitrogen atoms.  So, Let´s balance equations by inspection, just adding coefficients to each compound, until the amount of atoms are equal on both sides.   Step 1: 4NH3(g) + 5O2(g) ⟶4 NO(g) +6 H2O(l)  Step 2: 2NO(g) + O2(g) ⟶ 2NO2(g)  Step 3: 3NO2(g) + H2O(l) ⟶ 2HNO3(l) + NO(g)   Second we gather the information what we are going to use in our calculations. Final Mass of HNO3 = 7,839Kg  and Molecular Weigth = 63,01g/mol  NH3  Molecular Weight= 17,031 g/mol  Third we start discoverying the amount of NH3 that reacted completed to generate 7,839Kg of HNO3, using the giving equation and its respective molecular weights.  7,839Kg  of HNO3  x 1000g / 1Kg x 1mol HNO3 / 63,01g HNO3 x 3mol NO2 / 2 mol HNO3 x 2 mol NO/ 2 mol NO2 x 4 mol NH3/4 mol NO x 17,031g NH3/1 mol NH3 x 1 Kg / 1000g= 3,178 Kg NH3   The amount of NH3 that is required to produce 7,839 Kg of HNO3 is 3,178 Kg NH3