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

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:

$\text{Moles of ammonia}=\frac{56.8g}{17g/mol}=3.34mol$

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 = $\frac{5}{4}\times 3.34=4.175mol$ 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:

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

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