Q20E
Question
How many grams of oxygen gas are necessary to react completely with \({\bf{3}}{\bf{.01 \times 1}}{{\bf{0}}^{{\bf{21}}}}\) atoms of magnesium to yield magnesium oxide?
Step-by-Step Solution
VerifiedThe mass of oxygen is \(m\left( {{O_2}} \right) = 8.0 \times {10^{ - 2}}\;{\rm{kg}}\).
- Calcination converts both \(MgC{O_3}\)and \(Mg{\left( {OH} \right)_2}\)to \(MgO.\)
- The thermal treatment of the calcination process influences the surface area and pore size, and thus, the reactivity of the formed magnesium oxide.
- The level and nature of impurities present in the calcined material are heavily influenced by the source.
Start by converting the number of magnesium atoms to the number of moles of \(Mg.\)Convert the unit by dividing the number of atoms by Avogadro's number.
\(\begin{aligned}{l}n(Mg) = \frac{{N(Mg)}}{{{N_A}}}\\n(Mg) = \frac{{3.01 \times {{10}^{21}}}}{{6.022 \times {{10}^{23}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}}}\\n(Mg) = 5.0 \times {10^{ - 3}}\;{\rm{mol}}\end{aligned}\)
The reaction:
\(2Mg(s) + {O_2}(g) \to 2MgO(s)\)
To calculate the number of moles of \({O_2}\)produced, multiply the number of moles by the ratio of their stoichiometric coefficients (stoichiometric ratio).
\(\frac{{\nu (Mg)}}{{\nu \left( {{O_2}} \right)}} = \frac{2}{1}{\rm{ }}\)
\(\begin{aligned}{l}n\left( {{O_2}} \right) = n(Mg) \times \frac{{\nu \left( {{O_2}} \right)}}{{\nu (Mg)}}\\n\left( {C{l_2}} \right) = 5 \times {10^{ - 3}}\;{\rm{mol}} \times \frac{1}{2}\\n\left( {C{l_2}} \right) = 2.5 \times {10^{ - 3}}\;{\rm{mol}}\end{aligned}\)
Finally, calculate the mass of the oxygen.
\(\begin{aligned}{l}m\left( {{O_2}} \right) = n\left( {{O_2}} \right) \times M\left( {{O_2}} \right)\\m\left( {{O_2}} \right) = 2.5 \times {10^{ - 3}}\;{\rm{mol}} \times 32.0\;{\rm{g}}/{\rm{mol}}\\m\left( {{O_2}} \right) = 8.0 \times {10^{ - 2}}\;{\rm{kg}}\;\end{aligned}\)