Question#46528 – Chemistry – Inorganic Chemistry
Question:
The Ostwald process is used commercially to produce nitric acid, which is, in turn, used in many modern chemical processes. In the first step of the Ostwald process, ammonia is reacted with oxygen gas to produce nitric oxide and water. What is the maximum mass of H2O that can be produced by combining 86.4 g of each reactant?
4NH3(g)+5O2(g)→4NO(g)+6H2O(g)Answer:
Firstly, the amount of moles of each initial compound can be calculated:
n(NH3)=M(NH3)m(NH3)=17g/mol86.4g=5.1moln(O2)=M(O2)m(O2)=32g/mol86.4g=2.7mol
From the equation above it is obvious, that the theoretical molar ratio between NH3 and O2 is 4:5. In our case there is n(NH3):n(O2)=5.1:2.7.
Therefore, the limiting reactant is oxygen gas and ammonia is in excess. And we will continue calculations using amount of moles of O2.
If 5 moles of O2 produce 6 moles of water H2O, than 2.7 moles of O2 produce x moles of H2O. Using the following proportion, one can calculate x:
65=x2.7⇒x=52.7×6=3.24mol
The mass of water can be calculated considering that mass of one mole of water is 18 g/mol:
m(H2O)=n(H2O)×M(H2O)=3.24mol×18g/mol=58.32g
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