Chapter 5, Problem 5.77AP

(a)

Interpretation Introduction

Interpretation:

The number of grams of ${\text{CO}}_{\text{2}}$ are formed from $\text{2}\text{.5}$ moles of ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}$ has to be given.

Answer to Problem 5.77AP

The number of grams of ${\text{CO}}_{\text{2}}$ are formed from $\text{2}\text{.5}$ moles of ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}$ is $\text{220 g}\text{.}$

Explanation of Solution

Given,

Moles of ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}$ are $\text{2}\text{.5}$ moles.

The balanced chemical equation is,

  ${\text{2 C}}_{\text{2}}{\text{H}}_2{\text{+5 O}}_{\text{2}}\stackrel{\text{Δ}}{\to }{\text{4 CO}}_2\text{+2}{\text{H}}_{\text{2}}\text{O}$

The number of moles of carbon dioxide formed from $\text{2}\text{.5}$ moles of ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}$ is calculated as,

  $\text{2}\text{.5}\cancel{\text{mol}{\text{C}}_{\text{2}}{\text{H}}_2}\text{×}\frac{\text{4}\text{mol}{\text{CO}}_2}{\text{2}\cancel{\text{mol}{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}}}\text{=5}\text{mol}{\text{CO}}_2$

The number of moles of carbon dioxide formed from $\text{2}\text{.5}$ moles of ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}$ is $\text{5}\text{moles}\text{.}$

Moles can be converted into grams using the conversion factor:

  $\frac{\text{Molar}\text{mass}}{\text{1}\text{mol}}$

The number of grams present in $\text{2}\text{.5}$ moles of ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}$ is calculated as,

  $\text{5}\cancel{\text{mol}}\text{×}\frac{\text{44}\text{.01}\text{g}{\text{CO}}_{\text{2}}}{\text{1}\cancel{\text{mol}}}\text{=220}\text{.05}\text{g}$

The number of grams of ${\text{CO}}_{\text{2}}$ are formed from $\text{2}\text{.5}$ moles of ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}$ is $\text{220 g}\text{.}$

(b)

Interpretation Introduction

Interpretation:

The number of grams of ${\text{CO}}_{\text{2}}$ are formed from $\text{0}\text{.50}$ moles of ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}$ has to be given.

Answer to Problem 5.77AP

The number of grams of ${\text{CO}}_{\text{2}}$ are formed from $\text{0}\text{.50}$ moles of ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}$ is $\text{44 g}\text{.}$

Explanation of Solution

Given,

Moles of ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}$ are $\text{0}\text{.50}$ moles.

The balanced chemical equation is,

  ${\text{2 C}}_{\text{2}}{\text{H}}_2{\text{+5 O}}_{\text{2}}\stackrel{\text{Δ}}{\to }{\text{4 CO}}_2\text{+2}{\text{H}}_{\text{2}}\text{O}$

The number of moles of carbon dioxide formed from $\text{0}\text{.50}$ moles of ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}$ is calculated as,

  $\text{0}\text{.50}\cancel{\text{mol}{\text{C}}_{\text{2}}{\text{H}}_{\text{2}}}\text{×}\frac{\text{4}\text{mol}{\text{CO}}_{\text{2}}}{\text{2}\cancel{\text{mol}{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}}}\text{=1}\text{mol}{\text{CO}}_{\text{2}}$

The number of moles of carbon dioxide formed from $\text{0}\text{.50}$ moles of ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}$ is $\text{1}\text{mole}\text{.}$

Moles can be converted into grams using the conversion factor:

  $\frac{\text{Molar}\text{mass}}{\text{1}\text{mol}}$

The number of grams of ${\text{CO}}_{\text{2}}$ are formed from $\text{0}\text{.50}$ moles of ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}$ is calculated as,

  $\text{1}\cancel{\text{mol}}\text{×}\frac{\text{44}\text{.01}\text{g}{\text{CO}}_{\text{2}}}{\text{1}\cancel{\text{mol}}}\text{=44}\text{.01}\text{g}$

The number of grams of ${\text{CO}}_{\text{2}}$ are formed from $\text{0}\text{.50}$ moles of ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}$ is $\text{44 g}\text{.}$

(c)

Interpretation Introduction

Interpretation:

The number of grams of ${\text{H}}_{\text{2}}\text{O}$ are formed from $\text{0}\text{.25}$ moles of ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}$ has to be given.

Answer to Problem 5.77AP

The number of grams of ${\text{H}}_{\text{2}}\text{O}$ are formed from $\text{0}\text{.25}$ moles of ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}$ is $\text{4}\text{.5 g}\text{.}$

Explanation of Solution

Given,

Moles of ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}$ are $\text{0}\text{.25}$ moles.

The balanced chemical equation is,

  ${\text{2 C}}_{\text{2}}{\text{H}}_2{\text{+5 O}}_{\text{2}}\stackrel{\text{Δ}}{\to }{\text{4 CO}}_2\text{+2}{\text{H}}_{\text{2}}\text{O}$

The number of moles of water formed from $\text{0}\text{.25}$ moles of ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}$ is calculated as,

  $\text{0}\text{.25}\cancel{\text{mol}{\text{C}}_{\text{2}}{\text{H}}_{\text{2}}}\text{×}\frac{\text{2}\text{mol}{\text{H}}_{\text{2}}\text{O}}{\text{2}\cancel{\text{mol}{\text{C}}_{\text{2}}{\text{H}}_{\text{6}}}}\text{=0}\text{.25}\text{mol}{\text{H}}_{\text{2}}\text{O}$

The number of moles of water formed from $\text{0}\text{.25}$ moles of ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}$ is $\text{0}\text{.25 moles}\text{.}$

Moles can be converted into grams using the conversion factor:

  $\frac{\text{Molar}\text{mass}}{\text{1}\text{mol}}$

The number of grams of ${\text{H}}_{\text{2}}\text{O}$ are formed from $\text{0}\text{.25}$ moles of ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}$ is calculated as,

  $\text{0}\text{.25}\cancel{\text{mol}}\text{×}\frac{\text{18}\text{.02}\text{g}{\text{H}}_{\text{2}}\text{O}}{\text{1}\cancel{\text{mol}}}\text{=4}\text{.505}\text{g}$

The number of grams of ${\text{H}}_{\text{2}}\text{O}$ are formed from $\text{0}\text{.25}$ moles of ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}$ is $\text{4}\text{.5 g}\text{.}$