Answer is A - HCl is the titrant and D-NaOH is the analyte.
In a titration process, the solution of unknown concentration considered is the analyte. During titration, a standard solution(titrant) is added to an analyte until the equivalence point is achieved.
So in this case as the concentration of NaOH is not known,NaOH is the analyte.
Titrant is the solution whose concentration is known to us and is added to an analyte until the equivalence point is reached. In this case since we know the concentration of HCl, HCl is the titrant.
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
