99. How many grams of Epsom salts \(\left( {{\rm{MgS}}{{\rm{O}}_4} \cdot 7{{\rm{H}}_2}{\rm{O}}} \right)\) will form from\(5.0\;{\rm{kg}}\)of magnesium?

First,

\(n(Mg) = \frac{{m(Mg)}}{{M(Mg)}}\;\;\;\)

To calculate the number of moles of\(Mg\)that are in given mass of\(Mg\).

\(n(Mg) = \frac{{5.0 \cdot {{10}^3}g}}{{24.305\;{\rm{g}}/{\rm{mol}}}}\)

\(n(Mg) = 205.7\;{\rm{mol}}\).

\(m\left( {{\rm{MgS}}{{\rm{O}}_4} \cdot 7{{\rm{H}}_2}{\rm{O}}} \right) = n\left( {{\rm{MgS}}{{\rm{O}}_4} \cdot 7{{\rm{H}}_2}{\rm{O}}} \right) \cdot \)

Then, calculate the mass of Epson salt that has the same\(M\left( {{\rm{MgS}}{{\rm{O}}_4} \cdot 7{{\rm{H}}_2}{\rm{O}}} \right)\)number of moles as\(Mg\)that could be produced from it.

\(m\left( {{\rm{MgS}}{{\rm{O}}_4} \cdot 7{{\rm{H}}_2}{\rm{O}}} \right) = 205.7\;{\rm{mol}} \cdot 246.5\;{\rm{g}}/{\rm{mol}}\)

\(m\left( {MgS{O_4} \cdot 7{H_2}{\rm{O}}} \right) = 5.07 \cdot {10^4}g\).