he 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 93.3 g of each reactant?

148.2 g of H20

Explanation:

Equation of reaction: 4NH3 + 5O2 ---> 4NO + 6H20

From the equation above, 4 moles of ammonia reacts with 5 moles of oxygen gas to produce 6 moles water.

Molar mass of NH3 = 17 g/mol;

Molar mass of O2= 32 g/mol;

Molar mass of H2O = 18 g/mol

First, we determine the limiting reactant:

4*17 g of NH3 reacts with 5*32 g of O2

Mass Ratio = 68 : 160

Therefore, NH3 is the limiting reactant.

68 g of NH3 reacts to produce 6* 18 g of H20 = 108 g of H2O

93.3 g of NH3 will react to produce (93.3 * 108)/68 g of H20 = 148.2 g of H2O

Therefore, the maximum amount of H2O produced = 148.2 g

a saline water is fed to a reverse osmosis unit at flow rate o rotameter connected to reject (retentate) line reads 8 installed in reject and permeate line reads 15.2 milli siemens/cm and 300 micro siemens/com respectively.

answer : one can design an experiment using a liquid and an additive that can be added to the liquid and observe the chemical changes in the reaction species.

one would need a beaker which is filled with a known amount of the liquid in which thermometer is placed to measure the temperature changes; after addition of the additive to the liquid solution one can note down the temperature change and find whether the reaction is endothermic or exothermic in nature. by subtracting the initial temperature value with the final temperature one can find the temperature changes.

various combinations of liquid and additive can be used. liquid in this case can be a known quantity of water and additives can be epsom salt or without epsom salt the experiment can be conducted to notice the changes. water and ice, hydrogen peroxide and dry yeast, vinegar and baking soda,etc.