Chapter 5, Problem 5.78AP

(a)

Interpretation Introduction

Interpretation:

The number of grams of sodium hydrogen formed from $\text{3}\text{.0}$ moles of $\text{Na}$ has to be given.

Answer to Problem 5.78AP

The number of grams of sodium hydrogen formed from $\text{3}\text{.0}$ moles of $\text{Na}$ is $\text{120}\text{g}\text{.}$

Explanation of Solution

Given,

Moles of $\text{Na}$ are $\text{3}\text{.0}$ moles.

The balanced chemical equation is,

  $\text{2 Na}\left(\text{s}\right)\text{+2}{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\stackrel{\text{Δ}}{\to }\text{2}\text{NaOH}\left(\text{aq}\right){\text{+H}}_{\text{2}}\left(\text{g}\right)$

The number of moles of sodium hydroxide formed from $\text{3}\text{.0}$ moles of $\text{Na}$ is calculated as,

  $\text{3}\text{.0}\cancel{\text{mol}\text{Na}}\text{×}\frac{\text{2}\text{mol}\text{NaOH}}{\text{2}\cancel{\text{mol}\text{Na}}}\text{=3}\text{.0}\text{mol}\text{NaOH}$

The number of moles of sodium hydroxide formed from $\text{3}\text{.0}$ moles of $\text{Na}$ is $\text{3}\text{.0}\text{moles}\text{.}$

Moles can be converted into grams using the conversion factor:

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

The number of grams of sodium hydroxide formed from $\text{3}\text{.0}$ moles of $\text{Na}$ is calculated as,

  $\text{3}\cancel{\text{mol}}\text{×}\frac{\text{40}\text{g}\text{NaOH}}{\text{1}\cancel{\text{mol}}}\text{=120}\text{g}$

The number of grams of sodium hydrogen formed from $\text{3}\text{.0}$ moles of $\text{Na}$ is $\text{120}\text{g}\text{.}$

(b)

Interpretation Introduction

Interpretation:

The number of grams of hydrogen formed from $\text{0}\text{.30}$ moles of $\text{Na}$ has to be given.

Answer to Problem 5.78AP

The number of grams of hydrogen formed from $\text{0}\text{.30}$ moles of $\text{Na}$ is $\text{0}\text{.30g}\text{.}$

Explanation of Solution

Given,

Moles of $\text{Na}$ are $\text{0}\text{.30}$ moles.

The balanced chemical equation is,

  $\text{2 Na}\left(\text{s}\right)\text{+2}{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\stackrel{\text{Δ}}{\to }\text{2}\text{NaOH}\left(\text{aq}\right){\text{+H}}_{\text{2}}\left(\text{g}\right)$

The number of moles of hydrogen formed from $\text{0}\text{.30}$ moles of $\text{Na}$ is calculated as,

  $\text{0}\text{.30}\cancel{\text{mol}\text{Na}}\text{×}\frac{\text{1}\text{mol}{\text{H}}_{\text{2}}}{\text{2}\cancel{\text{mol}\text{Na}}}\text{=0}\text{.15mol}{\text{H}}_{\text{2}}$

The number of moles of hydrogen formed from $\text{0}\text{.30}$ moles of $\text{Na}$ is $\text{0}\text{.15}\text{moles}\text{.}$

Moles can be converted into grams using the conversion factor:

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

The number of grams of hydrogen formed from $\text{0}\text{.30}$ moles of $\text{Na}$ is calculated as,

  $\text{0}\text{.15}\cancel{\text{mol}}\text{×}\frac{\text{2}\text{.015}\text{g}{\text{H}}_{\text{2}}}{\text{1}\cancel{\text{mol}}}\text{=0}\text{.30}\text{g}$

The number of grams of hydrogen formed from $\text{0}\text{.30}$ moles of $\text{Na}$ is $\text{0}\text{.30g}\text{.}$

(c)

Interpretation Introduction

Interpretation:

The number of grams of water that is required to react with $\text{0}\text{.20}$ moles of $\text{Na}$ has to be given.

Answer to Problem 5.78AP

The number of grams of water that is required to react with $\text{0}\text{.20}$ moles of $\text{Na}$ is $\text{3}\text{.6}\text{g}\text{.}$

Explanation of Solution

Given,

Moles of $\text{Na}$ are $\text{0}\text{.20}$ moles.

The balanced chemical equation is,

  $\text{2 Na}\left(\text{s}\right)\text{+2}{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\stackrel{\text{Δ}}{\to }\text{2}\text{NaOH}\left(\text{aq}\right){\text{+H}}_{\text{2}}\left(\text{g}\right)$

The number of moles of water that is required to react with $\text{0}\text{.20}$ moles of $\text{Na}$ is calculated as,

  $\text{0}\text{.20}\cancel{\text{mol}\text{Na}}\text{×}\frac{\text{2}\text{mol}{\text{H}}_{\text{2}}\text{O}}{\text{2}\cancel{\text{mol}\text{Na}}}\text{=0}\text{.20}\text{mol}{\text{H}}_{\text{2}}\text{O}$

The number of moles of water that is required to react with $\text{0}\text{.20}$ moles of $\text{Na}$ is $\text{0}\text{.20}\text{moles}\text{.}$

Moles can be converted into grams using the conversion factor:

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

The number of grams of water that is required to react with $\text{0}\text{.20}$ moles of $\text{Na}$  is calculated as,

  $\text{0}\text{.20}\cancel{\text{mol}}\text{×}\frac{\text{18}\text{.02}\text{g}{\text{H}}_{\text{2}}\text{O}}{\text{1}\cancel{\text{mol}}}\text{=3}\text{.604}\text{g}$

The number of grams of water that is required to react with $\text{0}\text{.20}$ moles of $\text{Na}$ is $\text{3}\text{.6}\text{g}\text{.}$