Chapter 9, Problem 3RQ

(a)

Interpretation Introduction

Interpretation:

The statement that one mole of ${\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}$ produces 2 mole of ${\text{PH}}_{\text{3}}$ is correct or incorrect have to be given.

Answer to Problem 3RQ

The statement is correct.

Explanation of Solution

The given balanced equation is,

  ${\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}{\text{+6H}}_{\text{2}}\text{O}\stackrel{}{\to }\text{3Ca}{\left(\text{OH}\right)}_{\text{2}}{\text{+2PH}}_{\text{3}}$

In the mole ratio, the coefficients of the balanced equation are used.  Therefore the mole ratio is $\left(\frac{\text{2}\text{mol}{\text{PH}}_{\text{3}}}{\text{1}\text{mol}{\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}}\right)\text{.}$

The number of moles of phosphine can be calculated as,

  $\left(\text{1}\text{mol}{\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}\right)\left(\frac{\text{2}\text{mol}{\text{PH}}_{\text{3}}}{\text{1}\text{mol}{\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}}\right)\text{=2}\text{mol}{\text{PH}}_{\text{3}}$

The statement that one mole of ${\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}$ produces 2 mole of ${\text{PH}}_{\text{3}}$ is correct statement.

(b)

Interpretation Introduction

Interpretation:

The statement that one gram of ${\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}$ produces 2 gram of ${\text{PH}}_{\text{3}}$ is correct or incorrect has to be given.

Answer to Problem 3RQ

The statement is incorrect.

Explanation of Solution

Given,

The mass of ${\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}$ is $\text{1}\text{g}\text{.}$

The mass of phosphine is $\text{2}\text{g}\text{.}$

The molecular weight of phosphine is $\text{33}\text{.99}\text{g/mol}\text{.}$

The molecular weight of ${\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}$ is $\text{182}\text{.2}\text{g/mol}\text{.}$

The given balanced equation is,

  ${\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}{\text{+6H}}_{\text{2}}\text{O}\stackrel{}{\to }\text{3Ca}{\left(\text{OH}\right)}_{\text{2}}{\text{+2PH}}_{\text{3}}$

In the mole ratio, the coefficients of the balanced equation are used.  Therefore the mole ratio is $\left(\frac{\text{2}\text{mol}{\text{PH}}_{\text{3}}}{\text{1}\text{mol}{\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}}\right)\text{.}$

The number of grams of phosphine can be calculated as,

  $\text{1}\text{g}{\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}\left(\frac{\text{1}\text{mol}}{\text{182}\text{.2}\text{g}}\right)\left(\frac{\text{2}\text{mol}{\text{PH}}_{\text{3}}}{\text{1}\text{mol}{\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}}\right)\left(\frac{\text{33}\text{.99}\text{g}}{\text{1}\text{mol}}\right)\text{=0}\text{.373}{\text{PH}}_{\text{3}}$

The statement one gram of ${\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}$ produces 2 gram of ${\text{PH}}_{\text{3}}$ is an incorrect statement.

(c)

Interpretation Introduction

Interpretation:

The statement that three moles of $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ produced for each 2 mole of ${\text{PH}}_{\text{3}}$ is correct or incorrect has to be given.

Answer to Problem 3RQ

The statement is correct.

Explanation of Solution

The given balanced equation is,

  ${\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}{\text{+6H}}_{\text{2}}\text{O}\stackrel{}{\to }\text{3Ca}{\left(\text{OH}\right)}_{\text{2}}{\text{+2PH}}_{\text{3}}$

In the mole ratio, the coefficients of the balanced equation are used.  Therefore the mole ratio is $\left(\frac{\text{3}\text{mol}{\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}}{\text{2}\text{mol}{\text{PH}}_{\text{3}}}\right)\text{.}$

The statement that three moles of $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ produced for each 2 mole of ${\text{PH}}_{\text{3}}$ is correct statement.

(d)

Interpretation Introduction

Interpretation:

The statement that the mole ratio between phosphine and calcium phosphide is $\left(\frac{\text{2}\text{mol}{\text{PH}}_{\text{3}}}{\text{1}\text{mol}{\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}}\right)$ correct or incorrect has to be given.

Answer to Problem 3RQ

The statement is correct.

Explanation of Solution

The given balanced equation is,

  ${\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}{\text{+6H}}_{\text{2}}\text{O}\stackrel{}{\to }\text{3Ca}{\left(\text{OH}\right)}_{\text{2}}{\text{+2PH}}_{\text{3}}$

In the mole ratio, the coefficients of the balanced equation are used.  Therefore the mole ratio is $\left(\frac{\text{2}\text{mol}{\text{PH}}_{\text{3}}}{\text{1}\text{mol}{\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}}\right)\text{.}$

The statement is correct statement.

(e)

Interpretation Introduction

Interpretation:

The statement $2\text{mol}$ of ${\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}$ and $\text{3}\text{mol}$ of ${\text{H}}_{\text{2}}\text{O}$ reacts to form $4\text{mol}$ of phosphine is correct or incorrect has to be given.

Answer to Problem 3RQ

The statement is incorrect.

Explanation of Solution

The given balanced equation is,

  ${\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}{\text{+6H}}_{\text{2}}\text{O}\stackrel{}{\to }\text{3Ca}{\left(\text{OH}\right)}_{\text{2}}{\text{+2PH}}_{\text{3}}$

In the mole ratio, the coefficients of the balanced equation are used.  Therefore the mole ratio is $\left(\frac{\text{2}\text{mol}{\text{PH}}_{\text{3}}}{\text{6}{\text{mol H}}_{\text{2}}\text{O}}\right)\text{.}$

The number of moles of phosphine produced from water can be calculated as,

  $3{\text{mol H}}_{\text{2}}\text{O}\left(\frac{\text{2}\text{mol}{\text{PH}}_{\text{3}}}{\text{6}{\text{mol H}}_{\text{2}}\text{O}}\right)\text{=1}{\text{mol PH}}_{\text{3}}$

$2\text{mol}$ Of ${\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}$ will reacts with $\text{3}\text{mol}$ of ${\text{H}}_{\text{2}}\text{O}$ to produce $1\text{mol}$ of ${\text{PH}}_{\text{3}}\text{.}$  The statement is incorrect.

(f)

Interpretation Introduction

Interpretation:

The statement $2\text{mol}$ of ${\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}$ and $15\text{mol}$ of ${\text{H}}_{\text{2}}\text{O}$ reacts to form $6\text{mol}$ of $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ is correct or incorrect has to be given.

Answer to Problem 3RQ

The statement is correct.

Explanation of Solution

The given balanced equation is,

  ${\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}{\text{+6H}}_{\text{2}}\text{O}\stackrel{}{\to }\text{3Ca}{\left(\text{OH}\right)}_{\text{2}}{\text{+2PH}}_{\text{3}}$

In the mole ratio, the coefficients of the balanced equation are used.  Therefore the mole ratio is $\left(\frac{\text{3}\text{mol Ca}{\left(\text{OH}\right)}_{\text{2}}}{\text{1}\text{mol}{\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}}\right)\text{.}$

The number of moles of $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ produced from ${\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}$ can be calculated as,

  $\text{2}\text{mol}{\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}\left(\frac{\text{3}\text{mol Ca}{\left(\text{OH}\right)}_{\text{2}}}{\text{1}\text{mol}{\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}}\right)\text{=6}\text{mol}\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$

$2\text{mol}$ Of ${\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}$ will reacts with $12\text{mol}$ of ${\text{H}}_{\text{2}}\text{O}$ (3 moles in excess) to produce $6\text{mol}$ of $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}\text{.}$  The statement is correct.

(g)

Interpretation Introduction

Interpretation:

The possibility of limiting agent being ${\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}$ when $\text{200}\text{.0}\text{g}{\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}$ and $\text{100}\text{g}{\text{H}}_{\text{2}}\text{O}$ is reacted has to be given.

Concept Introduction:

The limiting reactant of the reaction is the reactant that is completely used during the reaction.  Using the mole ratio and starting amounts of the reactants limiting reactant can be determined.

Example:

Consider a reaction starts with $\text{30}{\text{g CaCO}}_{\text{3}}\text{\&}\text{11}\text{g}\text{HCl}\text{.}$  The values in grams has to be converted to moles by dividing with their molecular weights.  According to the mole ratio $\text{0}\text{.3}\text{g}$ of calcium carbonate require $\text{0}\text{.6}\text{g}$ of $\text{HCl}$ completely.  Therefore HCl is the limiting agent.

Answer to Problem 3RQ

The statement is incorrect.

Explanation of Solution

Given,

The mass of ${\text{H}}_{\text{2}}\text{O}$ is $\text{100}\text{g}\text{.}$

The mass of ${\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}$ is $\text{200}\text{g}\text{.}$

The molecular weight of phosphine is $\text{33}\text{.99}\text{g/mol}\text{.}$

The molecular weight of ${\text{H}}_{\text{2}}\text{O}$ is $\text{18}\text{.02}\text{g/mol}\text{.}$

The molecular weight of ${\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}$ is $\text{182}\text{.2}\text{g/mol}\text{.}$

The given balanced equation is,

  ${\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}{\text{+6H}}_{\text{2}}\text{O}\stackrel{}{\to }\text{3Ca}{\left(\text{OH}\right)}_{\text{2}}{\text{+2PH}}_{\text{3}}$

The amount of water needed to react with $\text{200}\text{g}$ of ${\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}$ can be calculated as,

  $\text{200}\text{g}{\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}\left(\frac{\text{1}\text{mol}}{\text{182}\text{.2}\text{g}}\right)\left(\frac{\text{6}\text{mol}{\text{H}}_{\text{2}}\text{O}}{\text{1}\text{mol}{\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}}\right)\left(\frac{\text{18}\text{.02}\text{g}}{\text{mol}}\right)\text{=119}\text{g}{\text{H}}_{\text{2}}\text{O}$

The amount of water present is less than needed to react with $\text{200}\text{g}$ of ${\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}.$  Water is the limiting agent.  The statement is incorrect.

(h)

Interpretation Introduction

Interpretation:

The theoretical yield of ${\text{PH}}_{\text{3}}$ is $\text{57}\text{.4}\text{g}$ correct or incorrect has to be given.

Answer to Problem 3RQ

The statement is incorrect.

Explanation of Solution

Given,

The mass of ${\text{H}}_{\text{2}}\text{O}$ is $\text{100}\text{g}\text{.}$

The mass of ${\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}$ is $\text{200}\text{g}\text{.}$

The molecular weight of phosphine is $\text{33}\text{.99}\text{g/mol}\text{.}$

The molecular weight of ${\text{H}}_{\text{2}}\text{O}$ is $\text{18}\text{.02}\text{g/mol}\text{.}$

The molecular weight of ${\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}$ is $\text{182}\text{.2}\text{g/mol}\text{.}$

The given balanced equation is,

  ${\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}{\text{+6H}}_{\text{2}}\text{O}\stackrel{}{\to }\text{3Ca}{\left(\text{OH}\right)}_{\text{2}}{\text{+2PH}}_{\text{3}}$

The theoretical yield of phosphine from $\text{100}\text{g}$ of water can be calculated by,

  $\text{100}\text{g}{\text{H}}_{\text{2}}\text{O}\left(\frac{\text{1}\text{mol}{\text{H}}_{\text{2}}\text{O}}{\text{18}\text{.02}\text{g}}\right)\left(\frac{\text{2}\text{mol}{\text{PH}}_{\text{3}}}{\text{6}\text{mol}{\text{H}}_{\text{2}}\text{O}}\right)\left(\frac{\text{33}\text{.99}\text{g}{\text{PH}}_{\text{3}}}{\text{1}\text{mol}}\right)\text{=62}\text{.9}\text{g}$

The theoretical yield of phosphine from $200\text{g}$ of ${\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}$ can be calculated by,

  $200\text{g}{\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}\left(\frac{\text{1}\text{mol}{\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}}{\text{182}\text{.2}\text{g}}\right)\left(\frac{\text{2}\text{mol}{\text{PH}}_{\text{3}}}{1\text{mol}{\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}}\right)\left(\frac{\text{33}\text{.99}\text{g}{\text{PH}}_{\text{3}}}{\text{1}\text{mol}}\right)\text{=74}\text{.62}\text{g}$

The theoretical yield of phosphine is $\text{62}\text{.9}\text{g}\text{\&74}\text{.62}\text{g}\text{.}$  The statement is incorrect.