Question

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?

Answer 1

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

Step-by-step explanation:

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

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}` .....(1)

Given mass of ammonia = 56.8 g

Molar mass of ammonia = 17 g/mol

Putting values in equation 1, we get:

`\text{Moles of ammonia}=(56.8g)/(17g/mol)=3.34mol`

For the given chemical equation:

`4NH_3+5O_2\rightarrow 4NO+6H_2O`

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 =
`(5)/(4)* 3.34=4.175mol` 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:

`4.175mol=\frac{\text{Mass of oxygen gas}}{32g/mol}\\\\\text{Mass of oxygen gas}=(4.175mol* 32g/mol)=133.6g`

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