Ammonia (nh3) reacts with oxygen to form nitric oxide (no) and water vapor: 4nh3 + 502 4no + 6h2o b) when 20.0 g nh3 and 50.0 g o2 are allowed to react, which is the limiting reagent? which reactant is left over and how much left over at the end of reaction? ans: 2.94 g c) what is the theoretical yield of no if 20.0 g nh3 and 50.0 g o2 are allowed to react? ans: 35.29 g d) what is the theoretical yield of h20 if 20.0 g nh3 and 50.0 g o2 are allowed to react? ans: 31.74 g

a) . Ammonia is the limiting reagent

b. Oxygen is left over and 0.1375 g of oxygen is left over.

c. The theoretical yield of is 35.29 g.

d. The theoretical yield of is 31.74 g.

Explanation:

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

For

Given mass of ammonia = 20.0 g

Molar mass of ammonia = 17.031 g/mol

Putting values in equation 1, we get:

For

Given mass of oxygen gas = 50.0 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

The chemical equation for the reaction is

By Stoichiometry of the reaction:

4 moles of ammonia reacts with = 5 moles of oxygen

So 1.17 moles of ammonia will react with = of oxygen

As, given amount of oxygen is more than the required amount. So, it is considered as an excess reagent and (1.6-1.4625)= 0.1375 g of oxygen is left unreacted.

Thus ammonia is considered as a limiting reagent because it limits the formation of product.

1. By Stoichiometry of the reaction:

4 moles of ammonia produces = 4 moles of

1.17 moles of ammonia will produce = of

Mass of

Thus Theoretical yield of is 35.29 grams.

2. By Stoichiometry of the reaction:

4 moles of ammonia produces = 6 moles of

1.2 moles of ammonia will produce = of

Mass of  

Thus Theoretical yield of is 31.74 grams.

your answer is a. the enthalpy of formation for a pure element in its standard state is always positive.