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.