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 51.0 g of each reactant? 4nh3(g)+5o2(g)⟶4no(g)+6h2o(g)

68.93 g of water

The equation for the reaction is;

4NH3(g)+5O2(g)⟶4NO(g)+6H2O(g)

We are given;

Mass of ammonia = 51.0 g

Mass of Oxygen = 51.0 g

We are required to determine the maximum amount of water that can be produced.

Moles = Mass ÷ Molar mass

Moles of ammonia gas

Molar mass of ammonia = 17.031 g/mol

Moles = 51.0 g ÷ 17.031 g/mol

          = 2.995 moles

          = 3 moles

Moles of Oxygen gas

Molar mass of oxygen = 16.0 g/mol

Moles = 51.0 g ÷ 16.0 g/mol

          = 3.1875 moles

But, from the reaction 4 moles of ammonia reacts with 5 moles of oxygen gas.

Therefore, Oxygen gas is the limiting reactant since the number of moles of oxygen gas are more compared to those of 3.1875 moles.

From the reaction, 5 moles of oxygen reacts to produce 6 moles of water.

Therefore, for 3.1875 moles of oxygen;

= 3.1875 moles × 6/5

= 3.825 moles of H₂O

Mass = number of moles × Molar mass

Molar mass of water = 18.02 g/mol

Therefore;

Moles of water = 3.825 moles × 18.02 g/mol

                         = 68.9265 g

                         = 68.93 g of water

Mass of water =  34.56 g

Explanation:

Given data:

Mass of ammonia = 51.0 g

Mass of oxygen = 51.0 g

Mass of water produced = ?

Chemical equation:

4NH₃ + 5O₂ → 4NO  + 6H₂O

Number of moles of oxygen:

Number of moles of oxygen = Mass /molar mass

Number of moles of oxygen = 51.0 g / 32 g/mol

Number of moles of oxygen = 1.6 mol

Number of moles of ammonia:

Number of moles of ammonia = Mass /molar mass

Number of moles of ammonia = 51.0 g / 17 g/mol

Number of moles of ammonia = 3 mol

Now we will compare the moles of water with ammonia and oxygen .

                NH₃            :               H₂O

                  4               :               6

                  3               :              6/4 ×3 = 4.5 mol

                 O₂              :                H₂O

                  5               :                6

                 1.6             :               6/5 ×  1.6= 1.92 mol

The number of moles of water produced by  oxygen are less it will be limiting reactant.

Mass of water = moles × molar mass

Mass of water = 1.92 mol × 18 g/mol

Mass of water =  34.56 g

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"