What mass in kg of NH3 must be used to produce 4.4 multiplied by 106 kg HNO3 by the Ostwald process, assuming 100% yield in each reaction?
This is a long problem to try to explain on this forum. Here is a site that shows you how to go through a simple stoichiometry problem (which this is) but you can just solve it more than once (or you can work out the proportions of NH3 initially to HNO3 in the end and just do the problem once).
http://www.jiskha.com/science/chemistry/stoichiometry.html
To find the mass of NH3 required to produce a given amount of HNO3, we need to use stoichiometry, which involves balancing the chemical equation and using the molar ratios.
The balanced chemical equation for the formation of HNO3 from NH3 using the Ostwald process is:
4 NH3 + 5 O2 → 4 NO + 6 H2O
2 NO + O2 → 2 NO2
3 NO2 + H2O → 2 HNO3 + NO
From the balanced equation, we can see that it takes 4 moles of NH3 to produce 4 moles of HNO3.
We need to calculate the moles of HNO3 first:
Given mass of HNO3 = 4.4 x 10^6 kg
To convert the mass of HNO3 to moles, we need the molar mass of HNO3:
Molar mass of HNO3 = 1 (H) + 14 (N) + 48 (3 O) = 63 g/mol = 0.063 kg/mol
Now we can calculate the moles of HNO3:
Moles of HNO3 = (given mass of HNO3) / (molar mass of HNO3)
= (4.4 x 10^6 kg) / (0.063 kg/mol)
Next, we can calculate the moles of NH3 required:
Moles of NH3 = Moles of HNO3 (by stoichiometry)
= 4 x (4.4 x 10^6 kg) / (0.063 kg/mol)
Finally, we can convert moles of NH3 to mass:
Mass of NH3 = Moles of NH3 x Molar mass of NH3
= Moles of NH3 x (Molar mass of N + Molar mass of H x 3)
= Moles of NH3 x (14 g/mol + 1 g/mol x 3)
= Moles of NH3 x 17 g/mol
Therefore, to calculate the mass of NH3 needed to produce 4.4 x 10^6 kg of HNO3 by the Ostwald process with 100% yield, you would perform the following calculation:
Mass of NH3 = (3.49 x 10^8 mol) x (17 g/mol)
= 5.92 x 10^9 g
= 5.92 x 10^6 kg
So, approximately 5.92 x 10^6 kg of NH3 would be required for the given reaction.