Given the following equation: |4 nh3 (g) + 5 о2 (g) —> 4 no (g) + 6 h20 () how many grams of h20 is produced if 21.1 grams of nh3 reacts with 73.9 grams of o2?

The mass of water that can be formed are 33.48 g.

Explanation:

To calculate the number of moles, we use the equation:

  ....(1)

Given mass of ammonia gas = 21.1 g

Molar mass of ammonia gas = 17 g/mol

Putting values in above equation, we get:  

Given mass of oxygen gas = 73.9 g

Molar mass of oxygen gas = 32 g/mol

Putting values in above equation, we get:  

For the given chemical equation:

By Stoichiometry of the reaction:

4 moles of ammonia gas reacts with 5 moles of oxygen gas.

So, 1.24 moles of ammonia gas will react with = of oxygen gas.

As, given amount of oxygen gas is more than the required amount. Thus, it is considered as an excess reagent.

So, ammonia gas is considered as a limiting reagent because it limits the formation of products.

By Stoichiometry of the above reaction:

4 moles of ammonia gas is producing 6 moles of water.

So, 1.24 moles of ammonia gas will produce = of water.

Now, calculating the mass of water by using equation 1, we get:

Moles of water = 1.86 moles

Molar mass of water = 18 g/mol

Putting all the values in equation 1, we get:

Hence, the mass of water that can be formed are 33.48 g