so you can see those fluorine atoms have really spread out around the central phosphorus atom. this gives us a trigonal bi-pyramidal molecular geometry for pf5.
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
