Question

MnO2 + 4HCl = Cl2 + MnCl2 + 2H20How many grams of MnCl2 will be produced when 4.75 moles of MnO2 react with excess HCl

Answer 1

Assuming all 4.75 MnO₂ will react, we have a stoichimetry of 1 to 1 of MnO₂ to MnCl₂, so we would produce the same number of moles of MnCl₂, that is:

`n_{MnCl_(2)}=n_{MnO_(2)}=4.75_{}`

To convert number of moles of MnCl₂ to grams of MnCl₂, we need the molecular weight of it, which can be calculated using the atomic mass of each of its atoms:

`M_(MnCl_2)=W_(Mn)+2\cdot W_(Cl)`

Consulting the atomic masses, we have:

`\begin{gathered} W_(Mn)\approx54.938g/mol \\ W_(Cl)\approx35.453g/mol \end{gathered}`

So:

`M_(MnCl_2)\approx(54.938+2\cdot35.453)g/mol=125.844g/mol`

Now, to find the mass, we have:

`\begin{gathered} M_{MnCl_(2)}=\frac{m_(MnCl_2)}{n_{MnCl_(2)}} \\ m_(MnCl_2)=M_(MnCl_2)\cdot n_(MnCl_2)=125.844g/mol\cdot4.75mol\approx598g \end{gathered}`

So, approximately 598 grams of MnCl₂ will be produced, assuming complete reaction.