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.

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