Question

Now we need to find the amount of NF3 that can be formed by the complete reactions of each of the reactants. If all of the N2 was used up in the reaction, how many moles of NF3 would be produced

Answer 1

The question is incomplete, the complete question is:

Nitrogen and fluorine react to form nitrogen fluoride according to the chemical equation:

$N_2(g)+3F_2(g)\rightarrow 2NF_3(g)$

A sample contains 19.3 g of $N_2$ is reacted with 19.3 g of $F_2$. Now we need to find the amount of $NF_3$ that can be formed by the complete reactions of each of the reactants.

If all of the $N_2$ was used up in the reaction, how many moles of $NF_3$ would be produced?

Answer: 1.378 moles of $NF_3$ are produced in the reaction.

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

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

Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.

Excess reagent is defined as the reagent which is left behind after the completion of the reaction.

In the given chemical reaction, $N_2$ is considered as a limiting reagent because it limits the formation of the product and it was completely consumed in the reaction.

We are given:

Mass of $N_2$ = 19.3 g

Molar mass of $N_2$ = 28.02 g/mol

Putting values in equation 1:

$\text{Moles of }N_2=\frac{19.3g}{28.02g/mol}=0.689mol$

For the given chemical reaction:

$N_2(g)+3F_2(g)\rightarrow 2NF_3(g)$

By the stoichiometry of the reaction:

1 mole of $N_2$ produces 2 moles of $NF_3$

So, 0.689 moles of $N_2$ will produce = $\frac{2}{1}\times 0.689=1.378mol$ of $NF_3$

Hence, 1.378 moles of $NF_3$ are produced in the reaction.