How many milliliters of 3.50 M HCl (aq) are required to react with 5.75 g of Zn(s)?

Question:

How many milliliters of 3.50 M HCl (aq) are required to react with 5.75 g of Zn(s)?

Stoichiometry

By making use of stoichiometric calculations, the reactant's amount required can be predicted in terms of mole. By using moles and the concentration of the reactants, its volume required can be calculated.

Answer and Explanation: 1


Given data:

  • The mass of {eq}{\rm{Zn}}\left( {\rm{s}} \right){/eq} is 5.75 g.
  • The concentration of {eq}{\rm{HCl}}{/eq} is 3.50 M.


The reaction of {eq}{\rm{Zn}}\left( {\rm{s}} \right){/eq} with {eq}{\rm{HCl}}{/eq} is shown below.

{eq}{\rm{Zn}}\left( {\rm{s}} \right) + 2{\rm{HCl}}\left( {{\rm{aq}}} \right) \to {\rm{ZnC}}{{\rm{l}}_2}\left( {{\rm{aq}}} \right) + {{\rm{H}}_{\rm{2}}}\left( {\rm{g}} \right){/eq}

First, calculate the number of moles of {eq}{\rm{Zn}}\left( {\rm{s}} \right){/eq} to do the stoichiometric calculations.

The moles of {eq}{\rm{Zn}}\left( {\rm{s}} \right){/eq} can be calculated by using the formula shown below.

{eq}n = \dfrac{m}{M}{/eq}

Where,

  • {eq}n{/eq} is the number of moles.
  • {eq}m{/eq} is the given mass.
  • {eq}M{/eq} is the molecular mass.

Molecular mass of {eq}{\rm{Zn}}\left( {\rm{s}} \right){/eq} is 65.38 g/mol.

Substitute the values in the above formula.

{eq}\begin{align*} n\left( {{\rm{Zn}}\left( {\rm{s}} \right)} \right) &= \dfrac{{5.75\;{\rm{g}}}}{{65.38\;{\rm{g/mol}}}}\\ &= 0.0879\;{\rm{mol}} \end{align*}{/eq}

It is seen in the reaction that,

1 mole of {eq}{\rm{Zn}}\left( {\rm{s}} \right){/eq} react with 2 moles of {eq}{\rm{HCl}}{/eq}.

Therefore, 0.0879 moles of {eq}{\rm{Zn}}\left( {\rm{s}} \right){/eq} react with {eq}2 \times 0.0879{/eq} moles of {eq}{\rm{HCl}}{/eq}.

Hence, the number of moles of {eq}{\rm{HCl}}{/eq} required is 0.1758.

The volume of {eq}{\rm{HCl}}{/eq} can be calculated by using the formula shown below.

{eq}{\rm{Volume}}\left( V \right) = \dfrac{{{\rm{Number}}\;{\rm{of}}\;{\rm{moles}}\left( n \right)}}{{{\rm{Concentration}}\left( M \right)}}{/eq}

Substitute the values in the above formula.

{eq}\begin{align*} {\rm{Volume}}\left( V \right) &= \dfrac{{{\rm{0}}{\rm{.1758}}\;{\rm{mol}}}}{{{\rm{3}}{\rm{.50}}\;{\rm{M}}}}\\ &= \dfrac{{{\rm{0}}{\rm{.1758}}\;{\rm{mol}}}}{{{\rm{3}}{\rm{.50}}\;{\rm{mol/L}}}}\\ &= 0.050{\rm{2}}\;{\rm{L}} \end{align*}{/eq}

The conversion of ml into L is shown below.

{eq}\begin{align*} 1\;{\rm{L}} &= {10^3}\;{\rm{mL}}\\ 0.0502\;{\rm{L}} &= 0.0502 \times {10^3}\;{\rm{mL}}\\ &= {\rm{50}}{\rm{.2}}\;{\rm{mL}} \end{align*}{/eq}

Hence, the volume in milliliters of {eq}{\rm{HCl}}{/eq} required is {eq}\boxed{{\rm{50}}{\rm{.2}}\;{\rm{mL}}}{/eq}


Learn more about this topic:

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Mole-to-Mole Ratios and Calculations of a Chemical Equation

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Chapter 9 / Lesson 2
147K

Learn about the mole ratio. Understand the definition of mole ratio, how to find mole ratio in stoichiometry, and see examples of using mole ratio in problems.


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