Chapter 9, Problem 13QAP

For each of the following balanced chemical equations, calculate how many grams of the product(s) would be produced by complete reaction of 0.125 mole of the first reactant.

msp;  ${\text{AgNO}}_3\left(aq\right)+\text{LiOH}\left(aq\right)\to \text{AgOH}\left(s\right)+{\text{LiNO}}_3\left(aq\right)$

msp;  ${\text{Al}}_2{\left({\text{SO}}_4\right)}_3\left(aq\right)+3{\text{CaCl}}_2\left(aq\right)\to 2{\text{AlCl}}_3\left(aq\right)+3{\text{CaSO}}_4\left(s\right)$

msp;  ${\text{CaCO}}_3\left(s\right)+2\text{HCl}\left(aq\right)\to {\text{CaCl}}_2\left(aq\right)+{\text{CO}}_2\left(g\right)+{\text{H}}_2\text{O}\left(l\right)$

msp;  $2{\text{C}}_4{\text{H}}_{10}\left(g\right)+13{\text{O}}_2\left(g\right)\to 8{\text{CO}}_2\left(g\right)+10{\text{H}}_2\text{O}\left(g\right)$

Interpretation Introduction

(a)

Interpretation:

Grams of the product(s) produced by complete reaction of $0.125$ mole of the first reactant in the given balanced chemical equation should be calculated.

Concept Introduction:

For a substance, number of moles is related to mass and molar mass of the substance as follows:

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

Here, m is mass of substance in g and M is molar mass of substance in g/mol.

Thus, from number of moles, mass can be calculated as follows:

$\text{m}=\text{n}\times \text{M}$.

Answer to Problem 13QAP

Mass of AgOH _{( s )} Produced = $15.6\text{ g}$

Mass of LiNO_{3 ( a q )} produced = $8.62\text{ g}$.

Explanation of Solution

According to the balanced equation, mole ratio between AgNO_{3 ( a q )} and AgOH _{( s )} = $\frac{{\text{1 mol AgOH}}_{\left(\text{s}\right)}}{{\text{1 mol AgNO}}_{\text{3 (aq)}}}$

So, Amount of AgOH _{( s )} produced = $0.125{\text{ mol AgNO}}_{\text{3}}{}_{\text{(aq)}}\times$

$\frac{{\text{1 mol AgOH}}_{\left(\text{s}\right)}}{{\text{1 mol AgNO}}_{\text{3 (aq)}}}$

= $0.12\text{5 mol}$

Molar mass of AgOH _{( s )} = $\left(107.9+16.00+1.028\right)\text{ g/mol}$ = $124.928\text{ g/ mol}$

Mass of AgOH _{( s )} Produced = $0.125\text{ mol }\times$

$124.92\text{8 g/ mol}$

= $15.6\text{ g}$

According to the balanced equation, mole ratio between AgNO_{3 ( a q )} and LiNO_{3 ( a q )} = $\frac{{\text{1 mol LiNO}}_{\text{3 (aq)}}}{{\text{1 mol AgNO}}_{\text{3 (aq)}}}$

So, Amount of LiNO_{3 ( a q )} produced = $0.125{\text{ mol AgNO}}_{\text{3}}{}_{\text{(aq)}}\times$

$\frac{{\text{1 mol LiNO}}_{\text{3 (aq)}}}{{\text{1 mol AgNO}}_{\text{3 (aq)}}}$

= $0.12\text{5 mol}$

Molar mass of LiNO_{3 ( a q )} = $\left\{(6.941+14.01+\left(3\times 16.00\right)\right\}\text{g/mol}$ = $68.95\text{1 g/mol}$

Mass of LiNO_{3 ( a q )} produced = $0.125\text{ mol ×}$

$68.9\text{51 g/mol}$

= $8.6\text{2 g}$.

Interpretation Introduction

(b)

Interpretation:

Grams of the product(s) produced by complete reaction of $0.125$ mole of the first reactant in the given balanced chemical equation should be calculated.

Concept Introduction:

For a substance, number of moles is related to mass and molar mass of the substance as follows:

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

Here, m is mass of substance in g and M is molar mass of substance in g/mol.

Thus, from number of moles, mass can be calculated as follows:

$\text{m}=\text{n}\times \text{M}$.

Answer to Problem 13QAP

Mass of AlCl_{3 ( a q )} produced = $33.3\text{ g}$

Mass of CaSO_{4 ( s )} produced = $51.1\text{ g}$.

Explanation of Solution

According to the balanced equation, mole ratio between Al₂ (SO₄ )_{3 ( a q )} and AlCl_{3 ( a q )} = $\frac{{\text{2 mol AlCl}}_{\text{3 (aq)}}}{{\text{1 mol Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3 (aq)}}}$

So, Amount of AlCl_{3 ( a q )} produced = $0.125{\text{ mol Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3 (aq)}}\times$

$\frac{{\text{2 mol AlCl}}_{\text{3 (aq)}}}{{\text{1 mol Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3 (aq)}}}$

= $0.250\text{ mol}$

Molar mass of AlCl_{3 ( a q )} = $\left\{26.98+\left(3\times 35.45\right)\right\}\text{ g/mol}$ = $133.33\text{ g/ mol}$

Mass of AlCl_{3 ( a q )} Produced = $0.250\text{ mol}$ × $133.33\text{ g/ mol}$

= $33.\text{3 g}$

According to the balanced equation, mole ratio between Al₂ (SO₄ )_{3 ( a q )} and CaSO_{4 ( s )} = $\frac{{\text{3 mol CaSO}}_{\text{4}}{}_{\text{(s)}}}{{\text{1 mol Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3 (aq)}}}$

So, Amount of CaSO_{4 ( s )} produced =

$\frac{{\text{3 mol CaSO}}_{\text{4}}{}_{\text{(s)}}}{{\text{1 mol Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3 (aq)}}}$

= $0.375\text{ mol}$

Molar mass of CaSO_{4 ( s )} = $\left\{40.08+32.07+\left(4\times 16.00\right)\right\}\text{ g/mol}$ = $\text{136}\text{.15 g/mol}$

Mass of CaSO_{4 ( s )} produced = $0.375\text{ mole}$ × $\text{136}\text{.15 g/mol}$

= $51.\text{1 g}$.

Interpretation Introduction

(c)

Interpretation:

Grams of the product(s) produced by complete reaction of $0.125$ mole of the first reactant in the given balanced chemical equation should be calculated.

Concept Introduction:

For a substance, number of moles is related to mass and molar mass of the substance as follows:

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

Here, m is mass of substance in g and M is molar mass of substance in g/mol.

Thus, from number of moles, mass can be calculated as follows:

$\text{m}=\text{n}\times \text{M}$.

Answer to Problem 13QAP

Mass of CaCl_{2 ( a q )} Produced = $13.9\text{ g}$

Mass of CO_{2 ( g )} produced = $5.5\text{0 g}$

Mass of H₂ O_{( l )} produced = $\text{2}\text{.23 g}$.

Explanation of Solution

According to the balanced equation, mole ratio between CaCO_{3 ( s )} and CaCl_{2 ( a q )} = $\frac{{\text{1 mol CaCl}}_{\text{2}}{}_{\left(\text{aq}\right)}}{{\text{1 mol CaCO}}_{\text{3 (s)}}}$

So, Amount of CaCl_{2 ( a q )} produced = $0.12{\text{5 mol CaCO}}_{\text{3 (s)}}\times$

$\frac{{\text{1 mol CaCl}}_{\text{2}}{}_{\left(\text{aq}\right)}}{{\text{1 mol CaCO}}_{\text{3 (s)}}}$

= $0.125\text{ mol}$

Molar mass of CaCl_{2 ( a q )} = $\left\{40.08+\left(2\times 35.45\right)\right\}\text{ g/mol}$ = $110.9\text{8 g/ mol}$

Mass of CaCl_{2 ( a q )} Produced = $0.125\text{ mol }\times$

$110.\text{98 g/ mol}$

= $\text{13}\text{.9 g}$

According to the balanced equation, mole ratio between CaCO_{3 ( s )} and CO_{2 ( g )} = $\frac{{\text{1 mol CO}}_{\text{2}}{}_{\left(\text{g}\right)}}{{\text{1 mol CaCO}}_{\text{3 (s)}}}$

So, Amount of CO_{2 ( g )} produced = $0.125{\text{ mol CaCO}}_{\text{3 (s)}}\times$

$\frac{{\text{1 mol CO}}_{\text{2}}{}_{\left(\text{g}\right)}}{{\text{1 mol CaCO}}_{\text{3 (s)}}}$

= $0.125\text{ mol}$

Molar mass of CO_{2 ( g )} = $\left\{12.01+\left(2\times 16.00\right)\right\}\text{ g/mol}$ = $\text{44}\text{.01 g/mol}$

Mass of CO_{2 ( g )} produced = $0.125\text{ mol }\times$

$\text{44}\text{.01 g/mol}$

= $5.50\text{ g}$

According to the balanced equation, mole ratio between CaCO_{3 ( s )} and H₂ O_{( l )} = $\frac{{\text{1 mol H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}}{{\text{1 mol CaCO}}_{\text{3 (s)}}}$

So, Amount of H₂ O_{( l )} produced = $0.125{\text{ mol CaCO}}_{3(s)}\times$

$\frac{{\text{1 mol H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}}{{\text{1 mol CaCO}}_{\text{3 (s)}}}$

= $0.125\text{ mol}$

Molar mass of H₂ O_{( l )} = $\left\{\left(2\times 1.008\right)+16.00\right\}\text{ g/mol}$ = $\text{18}\text{.016 }\text{g}/\text{m}\text{o}\text{l}$

Mass of H₂ O_{( l )} produced = $0.125\text{ mol }\times$

$\text{18}\text{.016 g/mol}$

= $\text{2}\text{.23 g}$.

Interpretation Introduction

(d)

Interpretation:

Grams of the product(s) produced by complete reaction of $0.125$ mole of the first reactant in the given balanced chemical equation should be calculated.

Concept Introduction:

For a substance, number of moles is related to mass and molar mass of the substance as follows:

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

Here, m is mass of substance in g and M is molar mass of substance in g/mol.

Thus, from number of moles, mass can be calculated as follows:

$\text{m}=\text{n}\times \text{M}$.

Answer to Problem 13QAP

Mass of CO_{2 ( g )} produced = $22.\text{0 g}$

Mass of H₂ O_{( g )} produced = $11.\text{3 g}$.

Explanation of Solution

According to the balanced equation, mole ratio between C₄ H_{1 0 ( g )} and CO_{2 ( g )} = $\frac{{\text{4 mol CO}}_{\text{2}}{}_{\left(\text{g}\right)}}{{\text{1 mol C}}_{\text{4}}{\text{H}}_{\text{10 (g)}}}$

So, Amount of CO_{2 ( g )} produced = $0.125{\text{ mol C}}_{\text{4}}{\text{H}}_{\text{10 (g)}}\times$

$\frac{{\text{4 mol CO}}_{\text{2}}{}_{\left(\text{g}\right)}}{{\text{1 mol C}}_{\text{4}}{\text{H}}_{\text{10 (g)}}}$

= $0.500\text{m}\text{o}\text{l}$

Molar mass of CO_{2 ( g )} = $\left\{12.01+\left(2\times 16.00\right)\right\}\text{ g/mol}$ = $\text{44}\text{.01 g/mol}$

Mass of CO_{2 ( g )} produced = $0.50\text{0 mol}$ × $\text{44}\text{.01 g/mol}$

= $22.0\text{ g}$

According to the balanced equation, mole ratio between C₄ H_{1 0 ( g )} and H₂ O_{( g )} = $\frac{{\text{5 mol H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}}{{\text{1 mol C}}_{\text{4}}{\text{H}}_{\text{10 (g)}}}$

So, Amount of H₂ O_{( g )} produced = $0.125{\text{ mol C}}_{\text{4}}{\text{H}}_{\text{10 (g)}}\text{×}$

$\frac{{\text{5 mol H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}}{{\text{1 mol C}}_{\text{4}}{\text{H}}_{\text{10 (g)}}}$

= $0.6\text{25 mol}$

Molar mass of H₂ O_{( g )} = $\left\{\left(2\times 1.008\right)+16.00\right\}\text{ g/mol}$ = $\text{18}\text{.016 g/mol}$

Mass of H₂ O_{( g )} produced = $0.62\text{5 mol}$ × $\text{18}\text{.016 g/mol}$

= $11.\text{3 g}$.