4 NH3 + 5 O2 > 4 NO + 6 H2O
How many moles and how many grams of oxygen (O2) are needed to react with 56.8 grams
of ammonia by this reaction?

The moles and mass of oxygen gas needed to react is 4.175 moles and 133.6 grams

Explanation:

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

    .....(1)

Given mass of ammonia = 56.8 g

Molar mass of ammonia = 17 g/mol

Putting values in equation 1, we get:

For the given chemical equation:

By Stoichiometry of the reaction:

4 moles of ammonia reacts with 5 mole of oxygen gas

So, 3.34 moles of ammonia will react with = of oxygen gas

Now, calculating the mass of oxygen gas from equation 1, we get:

Molar mass of oxygen gas = 32 g/mol

Moles of oxygen gas = 4.175 moles

Putting values in equation 1, we get:

Hence, the moles and mass of oxygen gas needed to react is 4.175 moles and 133.6 grams

the correct answer is:

a. formation of precipitation

it takes on an electronical energy