Q. 7.103 - An Introduction to General, Organic, and Biological Chemistry

Question

When ammonia (${\mathrm{NH}}_3$) gas reacts with fluorine gas, the gaseous products are dinitrogen tetrafluoride (${\mathrm{N}}_2{\mathrm{F}}_4$) and hydrogen fluoride ($\mathrm{HF}$). ($7.4, 7.7, 7.8$)

a. Write the balanced chemical equation.

b. How many moles of each reactant are needed to produce $4.00$ mole of $\mathrm{HF}$?

c. How many grams of ${\mathrm{F}}_2$ are needed to react with $25.5$ g of ${\mathrm{NH}}_3$?

d. How many grams of ${\mathrm{N}}_2{\mathrm{F}}_4$ can be produced when $3.40$ g of ${\mathrm{NH}}_3$ reacts?

Step-by-Step Solution

Answer

(a) The required balanced equation is $2{\mathrm{NH}}_3+5{\mathrm{F}}_2\to {\mathrm{N}}_2{\mathrm{F}}_4+6\mathrm{HF}$

(b) We require $1.33$ mole of ${\mathrm{NH}}_3$ and $3.33$ mole of ${\mathrm{F}}_2$

(c) The $142$ g of ${\mathrm{F}}_2$ are needed to react with $25.5$ g of ${\mathrm{NH}}_3$

(d) The $10.4$ g of ${\mathrm{N}}_2{\mathrm{F}}_4$ are needed to react with $3.40$ g of ${\mathrm{NH}}_3$

Part (a) Step 1: Given information

We need to find the balanced equation.

Part (a) Step 2: Explanation

We know that

The balanced equation is that in which the total mass of reactants is equal to the total mass of products.

Therefore, The balanced equation is, 

$2{\mathrm{NH}}_3+5{\mathrm{F}}_2\to {\mathrm{N}}_2{\mathrm{F}}_4+6\mathrm{HF}$

Part (b) Step 1: Given information

We need to find the no. of moles of each reactant are needed to produce $4.00$ mole of $\mathrm{HF}$

Part (b) Step 2: Explanation

From part (a)

We know that

The $6 \mathrm{mole}$of $\mathrm{HF}$is produced from $2$ mole of ${\mathrm{NH}}_3$ and $5$ mole of ${\mathrm{F}}_2$

Therefore, $4$ mole of $\mathrm{HF}$ is produced from $\frac{2 \mathrm{mole}}{6\cancel{ \mathrm{mole}}}\times 4 \cancel{\mathrm{mole}}=1.33 \mathrm{mole}$ of ${\mathrm{NH}}_3$ and $\frac{5 \mathrm{mole}}{6 \cancel{\mathrm{mole}}}\times 4\cancel{ \mathrm{mole}}= 3.33 \mathrm{mole}$ of ${\mathrm{F}}_2$

Part (c) Step 1: Given information

We need to find the mass of ${\mathrm{F}}_2$ needed to react with $25.5$ g of ${\mathrm{NH}}_3$

Part (c) Step 2: Explanation

From part (a)

We know that 

The $34$ g of ${\mathrm{NH}}_3$reacts with $190$ g of ${\mathrm{F}}_2$

Therefore, $25.5$ g of ${\mathrm{NH}}_3$ reacts with $\frac{190 \cancel{\mathrm{g}}}{34 \cancel{\mathrm{g}}}\times 25.5 \mathrm{g}=142 \mathrm{g}$

Part (d) Step 1: given information

We need to find the mass of ${\mathrm{N}}_2{\mathrm{F}}_4$ produced when $3.40$ g of ${\mathrm{NH}}_3$ reacts

Part (d) Step 2: Explanation

From part (a) 

We know that

The $34$ g of ${\mathrm{NH}}_3$ produces $104$ g of ${\mathrm{N}}_2{\mathrm{F}}_4$

Therefore, $3.40$ g of ${\mathrm{NH}}_3$ produces $\frac{104 \mathrm{g}}{34 \cancel{\mathrm{g}}}\times 3.4 \cancel{\mathrm{g}}=10.4 \mathrm{g}$ of ${\mathrm{N}}_2{\mathrm{F}}_4$