Chapter 3, Problem 74QAP

Magnesium ribbon reacts with acid to produce hydro- gen gas and magnesium ions. Different masses of magnesium ribbon are added to 10 mL of the acid. The volume of the hydrogen gas obtained is a measure of the number of moles of hydrogen produced by the reaction. Various measurements are given in the table below.

(a) Draw a graph of the results by plotting the mass of Mg versus the volume of the hydrogen gas.

(b) What is the limiting reactant in experiment 1?

(c) What is the limiting reactant in experiment 3?

(d) What is the limiting reactant in experiment 6?

(e) Which experiment uses stoichiometric amounts of each reactant?

(f) What volume of gas would be obtained if 0.300 g of Mg ribbon were used? If 0.010 g were used?

Interpretation Introduction

(a)

Interpretation:.

The graph between mass of Mg versus the volume of hydrogen gas should be plotted..

Concept introduction:.

The number of moles of a substance is related to mass and molar mass as follows:.

$n=\frac{m}{M}$.

Here,mis mass andMis molar mass of the substance..

The density of solution can be calculated as follow:.

$d=\frac{m}{V}$.

Here, m is mass and V is volume.

Answer to Problem 74QAP

Explanation of Solution

The data of mass of Mg ribbon in grams and volume of hydrogen gas produced in experiments is as follows:.

Experiment	Mass of Mg ribbon (g)	Volume of acid used (mL)	Volume of hydrogen gas (mL)
1	0.020	10.0	21
2	0.040	10.0	42
3	0.080	10.0	82
4	0.120	10.0	122
5	0.160	10.0	122
6	0.200	10.0	122

.

To plot put the data of mass of Mg ribbon on x-axis and volume of hydrogen gas at y-axis:.

Interpretation Introduction

(b)

Interpretation:

The limiting reactant in experiment 1 should be determined..

Concept introduction:.

The number of moles of a substance is related to mass and molar mass as follows:.

$n=\frac{m}{M}$.

Here,mis mass andMis molar mass of the substance..

The density of solution can be calculated as follow:.

$d=\frac{m}{V}$.

Here, m is mass and V is volume.

Answer to Problem 74QAP

Mg is limiting reactant.

Explanation of Solution

The balanced chemical reaction will be as follows:.

$\text{Mg}\left(s\right)+2\text{HA}\left(aq\right)\to {\text{MgA}}_{\text{2}}\left(aq\right)+{\text{H}}_{\text{2}}\left(g\right)$.

According to experiment 1, mass of Mg ribbon is 0.020 g, volume of acid used is 10.0 mL and volume of ${\text{H}}_{\text{2}}$ gas is 21 mL..

The density of ${\text{H}}_{\text{2}}$ gas is 0.08988 g/L. The mass of ${\text{H}}_{\text{2}}$ gas can be calculated as follows:.

$m=d\times V$.

Putting the values,

$m=\left(0.08988\text{ g/L}\right)\left(21\text{ mL}\right)\left(\frac{{10}^{-3}\text{ L}}{1\text{ mL}}\right)=1.8875\times {10}^{-3}\text{ g}$.

Molar mass of ${\text{H}}_{\text{2}}$ is 2.016 g/mol thus, number of moles of ${\text{H}}_{\text{2}}$ will be:.

$n=\frac{1.8875\times {10}^{-3}\text{ g}}{2.016\text{ g/mol}}=9.36\times {10}^{-4}\text{ mol}$.

From the balanced chemical reaction, 1 mol of hydrogen gas is produced from 1 mol of Mg thus, number of moles of Mg required to produce $9.36\times {10}^{-4}\text{ mol}$ of hydrogen gas will be $9.36\times {10}^{-4}\text{ mol}$..

The mass of Mg is 0.020 g and molar mass of Mg is 24.305 g/mol thus, number of moles of Mg will be:.

$n=\frac{m}{M}=\frac{0.020\text{ g}}{24.305\text{ g/mol}}=8.23\times {\text{10}}^{-4}\text{ mol}$.

Since, number of moles of Mg required is $9.36\times {10}^{-4}\text{ mol}$ thus, Mg is limiting reactant in experiment 1.

Interpretation Introduction

(c)

Interpretation:

The limiting reactant in experiment 3 should be determined..

Concept introduction:.

The number of moles of a substance is related to mass and molar mass as follows:.

$n=\frac{m}{M}$.

Here,mis mass andMis molar mass of the substance..

The density of solution can be calculated as follow:.

$d=\frac{m}{V}$.

Here, m is mass and V is volume.

Answer to Problem 74QAP

Mg is limiting reactant.

Explanation of Solution

The balanced chemical reaction will be as follows:.

$\text{Mg}\left(s\right)+2\text{HA}\left(aq\right)\to {\text{MgA}}_{\text{2}}\left(aq\right)+{\text{H}}_{\text{2}}\left(g\right)$.

According to experiment 3, mass of Mg ribbon is 0.080 g, volume of acid used is 10.0 mL and volume of ${\text{H}}_{\text{2}}$ gas is 82 mL..

The density of ${\text{H}}_{\text{2}}$ gas is 0.08988 g/L. The mass of ${\text{H}}_{\text{2}}$ gas can be calculated as follows:.

$m=d\times V$.

Putting the values,

$m=\left(0.08988\text{ g/L}\right)\left(82\text{ mL}\right)\left(\frac{{10}^{-3}\text{ L}}{1\text{ mL}}\right)=7.37\times {10}^{-3}\text{ g}$.

Molar mass of ${\text{H}}_{\text{2}}$ is 2.016 g/mol thus, number of moles of ${\text{H}}_{\text{2}}$ will be:.

$n=\frac{7.37\times {10}^{-3}\text{ g}}{2.016\text{ g/mol}}=3.66\times {10}^{-3}\text{ mol}$.

From the balanced chemical reaction, 1 mol of hydrogen gas is produced from 1 mol of Mg thus, number of moles of Mg required to produce $3.66\times {10}^{-3}\text{ mol}$ of hydrogen gas will be $3.66\times {10}^{-3}\text{ mol}$..

The mass of Mg is 0.080 g and molar mass of Mg is 24.305 g/mol thus, number of moles of Mg will be:.

$n=\frac{m}{M}=\frac{0.080\text{ g}}{24.305\text{ g/mol}}=3.29\times {\text{10}}^{-3}\text{ mol}$.

Since, number of moles of Mg required is $3.66\times {10}^{-3}\text{ mol}$ thus, Mg is limiting reactant in experiment 3.

Interpretation Introduction

(d)

Interpretation:

The limiting reactant in experiment 6 should be determined..

Concept introduction:.

The number of moles of a substance is related to mass and molar mass as follows:.

$n=\frac{m}{M}$.

Here,mis mass andMis molar mass of the substance..

The density of solution can be calculated as follow:.

$d=\frac{m}{V}$.

Here, m is mass and V is volume.

Answer to Problem 74QAP

Acid is limiting reactant.

Explanation of Solution

The balanced chemical reaction will be as follows:.

$\text{Mg}\left(s\right)+2\text{HA}\left(aq\right)\to {\text{MgA}}_{\text{2}}\left(aq\right)+{\text{H}}_{\text{2}}\left(g\right)$.

According to experiment 6, mass of Mg ribbon is 0.200 g, volume of acid used is 10.0 mL and volume of ${\text{H}}_{\text{2}}$ gas is 122 mL..

The density of ${\text{H}}_{\text{2}}$ gas is 0.08988 g/L. The mass of ${\text{H}}_{\text{2}}$ gas can be calculated as follows:.

$m=d\times V$.

Putting the values,

$m=\left(0.08988\text{ g/L}\right)\left(122\text{ mL}\right)\left(\frac{{10}^{-3}\text{ L}}{1\text{ mL}}\right)=0.011\text{ g}$.

Molar mass of ${\text{H}}_{\text{2}}$ is 2.016 g/mol thus, number of moles of ${\text{H}}_{\text{2}}$ will be:.

$n=\frac{\text{0}\text{.011 g}}{2.016\text{ g/mol}}=5.44\times {10}^{-3}\text{ mol}$.

From the balanced chemical reaction, 1 mol of hydrogen gas is produced from 1 mol of Mg thus, number of moles of Mg required to produce $5.44\times {10}^{-3}\text{ mol}$ of hydrogen gas will be $5.44\times {10}^{-3}\text{ mol}$..

The mass of Mg is 0.200 g and molar mass of Mg is 24.305 g/mol thus, number of moles of Mg will be:.

$n=\frac{m}{M}=\frac{0.200\text{ g}}{24.305\text{ g/mol}}=8.23\times {\text{10}}^{-3}\text{ mol}$.

Since, number of moles of Mg required is $5.44\times {10}^{-3}\text{ mol}$ thus, Mg is in excess in experiment 6 and acid is limiting reactant.

Interpretation Introduction

(e)

Interpretation:

The experiment that uses stoichiometric amounts of each reactant should be determined..

Concept introduction:.

The number of moles of a substance is related to mass and molar mass as follows:.

$n=\frac{m}{M}$.

Here,mis mass andMis molar mass of the substance..

The density of solution can be calculated as follow:.

$d=\frac{m}{V}$.

Here, m is mass and V is volume.

Answer to Problem 74QAP

Experiment 4.

Explanation of Solution

According to balance chemical reaction, 1 mol of Mg gives 1 mol of hydrogen gas thus, the experiment in which same number of moles of Mg reacts with acid to form hydrogen gas that experiment uses stoichiometric amounts of each reactant..

This cannot be experiment 1, 3 and 6 because ratio of number of moles of Mg and hydrogen gas is not 1:1 in these experiments..

Check experiment 2: mass of Mg is 0.040 g and molar mass of Mg is 24.305 g/mol thus, number of mol of Mg will be:

$n=\frac{0.040\text{ g}}{24.305\text{ g/mol}}=1.65\times {10}^{-3}\text{ mol}$.

The volume of ${\text{H}}_{\text{2}}$ gas is 42 mL..

The density of ${\text{H}}_{\text{2}}$ gas is 0.08988 g/L. The mass of ${\text{H}}_{\text{2}}$ gas can be calculated as follows:.

$m=d\times V$.

Putting the values,

$m=\left(0.08988\text{ g/L}\right)\left(42\text{ mL}\right)\left(\frac{{10}^{-3}\text{ L}}{1\text{ mL}}\right)=3.77\times {10}^{-3}\text{ g}$.

Molar mass of ${\text{H}}_{\text{2}}$ is 2.016 g/mol thus, number of moles of ${\text{H}}_{\text{2}}$ will be:.

$n=\frac{3.77\times {10}^{-3}\text{ g}}{2.016\text{ g/mol}}=1.87\times {10}^{-3}\text{ mol}$.

The number of moles of Mg and hydrogen gas is not same thus, it is not experiment 2..

Check experiment 4: mass of Mg is 0.120 g and molar mass of Mg is 24.305 g/mol thus, number of mol of Mg will be:

$n=\frac{0.120\text{ g}}{24.305\text{ g/mol}}=5.0\times {10}^{-3}\text{ mol}$.

The volume of ${\text{H}}_{\text{2}}$ gas is 122 mL..

The density of ${\text{H}}_{\text{2}}$ gas is 0.08988 g/L. The mass of ${\text{H}}_{\text{2}}$ gas can be calculated as follows:.

$m=d\times V$.

Putting the values,

$m=\left(0.08988\text{ g/L}\right)\left(122\text{ mL}\right)\left(\frac{{10}^{-3}\text{ L}}{1\text{ mL}}\right)=0.011\text{ g}$.

Molar mass of ${\text{H}}_{\text{2}}$ is 2.016 g/mol thus, number of moles of ${\text{H}}_{\text{2}}$ will be:.

$n=\frac{0.011\text{ g}}{2.016\text{ g/mol}}=5.4\times {10}^{-3}\text{ mol}$.

The number of moles of Mg and hydrogen gas is approximately same thus, it is experiment 4..

Check experiment 5: mass of Mg is 0.160 g and molar mass of Mg is 24.305 g/mol thus, number of mol of Mg will be:

$n=\frac{0.160\text{ g}}{24.305\text{ g/mol}}=6.58\times {10}^{-3}\text{ mol}$.

The volume of ${\text{H}}_{\text{2}}$ gas is 122 mL..

The density of ${\text{H}}_{\text{2}}$ gas is 0.08988 g/L. The mass of ${\text{H}}_{\text{2}}$ gas can be calculated as follows:.

$m=d\times V$.

Putting the values,

$m=\left(0.08988\text{ g/L}\right)\left(122\text{ mL}\right)\left(\frac{{10}^{-3}\text{ L}}{1\text{ mL}}\right)=0.011\text{ g}$.

Molar mass of ${\text{H}}_{\text{2}}$ is 2.016 g/mol thus, number of moles of ${\text{H}}_{\text{2}}$ will be:.

$n=\frac{0.011\text{ g}}{2.016\text{ g/mol}}=5.4\times {10}^{-3}\text{ mol}$.

The number of moles of Mg and hydrogen gas is not same thus, it is not experiment 4..

Therefore, experiment 4 uses stoichiometric amounts of each reactant.

Interpretation Introduction

(f)

Interpretation:

The volume of the gas for 0.300 g and 0.010 g of Mg ribbon should be calculated.

Concept introduction:.

The number of moles of a substance is related to mass and molar mass as follows:.

$n=\frac{m}{M}$.

Here,mis mass andMis molar mass of the substance..

The density of solution can be calculated as follow:.

$d=\frac{m}{V}$.

Here, m is mass and V is volume.

Answer to Problem 74QAP

The volume of hydrogen gas produced from 0.120 g of Mg and 0.010 g of Mg is 122 mL and 11.32 mL respectively.

Explanation of Solution

The graph between mass of Mg ribbon and volume of hydrogen gas is as follows:.

According to the graph, above the mass of Mg 0.120 g, the volume of hydrogen gas becomes constant at 122 mL thus, the volume of hydrogen gas produced if 0.120 g of Mg is burned will be 122 mL.

Considering only the straight line in the graph,.

Experiment	Mass of Mg ribbon (g)	Volume of acid used (mL)	Volume of hydrogen gas (mL)
1	0.020	10.0	21
2	0.040	10.0	42
3	0.080	10.0	82
4	0.120	10.0	122

.

The plot will be as follows:.

Comparing this with equation of straight line $y=mx+c$ here, m is slope and c is intercept thus, 1007 is slope and 1.254 is intercept.

For the mass of ribbon 0.010 g, the volume of hydrogen gas can be calculated as follows:.

$y=\left(1007\right)\left(0.010\right)+1.254=11.32\text{ mL}$.

Therefore, the volume of hydrogen gas is 11.32 mL.