Chapter 16, Problem 77A

Interpretation Introduction

Interpretation:

To determine the volume of hydrogen gas produced at STP during the 3.00 min reactions.

Concept introduction:

In a chemical reaction, the reactants are written on the left side of the equation and the product is written on the right side.

The average rate of a reaction can be given by the formula,

Average reaction rate $\text{=}-\frac{\text{Δ}\left[\text{reactant}\right]}{\text{Δt}}$

Here, $\text{Δ}\left[\text{reactant}\right]$ = change in the concentration of the reactant

Answer to Problem 77A

The volume of hydrogen gas produced at STP during the 3.00 min reactions is 1.38 L.

Explanation of Solution

Given chemical reaction,

$\text{Mg}\left(\text{s}\right)\text{+}\text{2HCl}\left(\text{aq}\right)\to {\text{H}}_{\text{2}}\left(\text{g}\right)\text{+}{\text{MgCl}}_{\text{2}}\left(\text{aq}\right)$. In this reaction, the mass of a sample of magnesium is obtained and the sample is placed in a container of hydrochloric acid.

Also given the value of mass of Mg and Volume of H and time in table :

Time (min)	Mass of Mg (g)	Volume of H at STP (L)
0.00	6.00	0.00
3.00	4.50	?

From table, one can see that the mass of Mg decreases from 6.00 g at 0.00 min to 4.50 g at 3.00 min. So, the change in time is $\Delta t=3.00\min -0.00\min =3.00\min$ and the change in mass is $\text{4}\text{.50}\text{g}-\text{6}\text{.00}\text{g}\text{=}-\text{1}\text{.50}\text{g}$.

Now, the mass of Mg is converted into moles by multiplying it with the conversion factor for the molar mass. Here, $\text{1}\text{mol=}\text{24}\text{.3}\text{g}$

Moles of Mg = $\left(-\text{1}\text{.50}\text{g}\right)\left(\frac{\text{mol}}{\text{24}\text{.3}\text{g}}\right)$

$\begin{array}{l}=\left(\frac{-\text{1}\text{.50}}{\text{ 24}\text{.3}}\right)\text{mol}\\ \text{=}-\text{0}\text{.6017}\text{mol}\end{array}$

Therefore, the change in the number of moles per liter of Mg is $-\text{0}\text{.6017}\text{mol}$

Using these values now we can calculate the rate of the reaction using the following reaction,

Average reaction rate $\text{=}-\frac{\text{Δ}\left[\text{reactant}\right]}{\text{Δt}}$

$\begin{array}{l}\text{=}-\frac{-\text{0}\text{.6017}\text{mol}}{\text{3}\text{.00}\text{min}}\\ \text{=}\frac{\text{0}\text{.6017}}{\text{3}\text{.00}}\frac{\text{mol}}{\text{min}}\\ \text{=}\text{0}\text{.0206}\frac{\text{mol}}{\text{min}}\end{array}$

Therefore, the average reaction rate is $\text{0}\text{.0206}\frac{\text{mol}}{\text{min}}$

Now, to calculate the moles of hydrogen gas left after 3 minutes, we multiply the reaction rate by 3.00 minutes.

$\begin{array}{c}\text{Moles of}{\text{H}}_{\text{2}}\text{=}\left(\text{0}\text{.0206}\frac{\text{mol}}{\text{min}}\right)\left(3.00\min \right)\\ =0.0618\text{mol}\end{array}$

We know that 1 mole of an ideal gas such as hydrogen occupies 22.4 L at STP. So, to determine the volume of hydrogen gas after 3 minutes we multiply the number of moles 0.0618 mol by the factor $\left(\frac{\text{22}\text{.4}\text{L}}{\text{1}\text{mol}}\right)$.

$\begin{array}{c}\text{Volume}\text{of}{\text{H}}_{\text{2}}\text{=}(0.0618\text{mol)}\left(\frac{\text{22}\text{.4}\text{L}}{\text{1}\text{mol}}\right)\\ =(0.0618\times \text{22}\text{.4)}\text{L}\\ \text{=1}\text{.38}\text{L}\end{array}$

Therefore, the volume of hydrogen gas that was produced during the reaction is 1.38 L.

Conclusion

The volume of hydrogen gas that was produced during the reaction is 1.38 Lat STP during the 3.00 min reactions.