How many grams of hydrogen are produced if you have 120 grams of sodium? 2 Na + 2 H_2O...

Question:

How many grams of hydrogen are produced if you have {eq}120 \ grams {/eq} of sodium ?

{eq}2 Na + 2 H_2O \longrightarrow 2 NaOH + H_2. {/eq}

Stoichiometry

The stoichiometry of any reaction is based upon a law which is called conservation of mass. According to this law, the overall mass of reactants is equal to the overall mass of products.

Answer and Explanation:

Given data

The given weight of sodium is 120 g.

The molecular weight of sodium is 22.989 g/mol.


The formula to calculate number of moles is given below.

{eq}n = \dfrac{m}{M}......({\rm{I}}) {/eq}


Where,

  • N is the number of moles
  • m is the given weight.
  • M is the molecular weight.


Substitute the values of n and M in expression(I).

{eq}\begin{align*} n &= \dfrac{{{\rm{120}}\;{\rm{g}}}}{{{\rm{22}}{\rm{.98}}\;{\rm{g/mol}}}}\\ n &= 5.221\;{\rm{mol}} \end{align*} {/eq}


The reaction of sodium with water is shown below.


{eq}{\rm{2Na}} + {\rm{2}}{{\rm{H}}_{\rm{2}}}{\rm{O}} \to {\rm{2NaOH}} + {{\rm{H}}_{\rm{2}}} {/eq}


Here, two moles of sodium reacts with two moles of water to give two moles of NaOH and one mole of {eq}{{\rm{H}}_{\rm{2}}} {/eq}.


The relation between the number of moles of sodium and number of moles of hydrogen is given below.


{eq}\dfrac{{{n_{Na}}}}{2} = \dfrac{{{n_{{H_2}}}}}{1} {/eq}


Substitute the value of n in above expression.

{eq}\begin{align*} \dfrac{{5.221\;{\rm{mol}}}}{2} &= \dfrac{{{n_{{H_2}}}}}{1}\\ {n_{{H_2}}} &= \dfrac{{5.221}}{2}\\ {n_{{H_2}}} &= 2.61\;{\rm{mol}} \end{align*} {/eq}


The molecular weight of {eq}{{\rm{H}}_{\rm{2}}} {/eq} is 2.015g/mol.


Substitute the value of n and M in expression I.


{eq}\begin{align*} 2.61 &= \dfrac{m}{{2.01}}\\ m &= {\rm{5}}{\rm{.246}}\;{\rm{g}} \end{align*} {/eq}


Therefore, the mass of {eq}{{\rm{H}}_{\rm{2}}} {/eq} when 120g of sodium is present is 5.246g


Learn more about this topic:

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