Mass of MgO = 28.35grams
Mass of Sulphur = 11.29 grams
The balanced chemical equation between magnesium metal and sulphur dioxide is given as:
[tex]2Mg+SO_2\rightarrow2MgO+S[/tex]Determine the moles of magnesium
Mole = mass/molar mass
Mole of Mg = 17.1/24.305
mole of Mg = 0.704moles
According to stoichiometry, 2 moles of Mg produces 2 moles of MgO, hence the required mass of MgO will be:
[tex]\begin{gathered} Mass\text{ of MgO}=0.704\times40.3 \\ Mass\text{ of MgO}=28.35grams \end{gathered}[/tex]Similarly, 2moles of Mg produces 1 mole of sulphur, hence the mass of sulphur produced is;
[tex]\begin{gathered} Mass\text{ of S}=\frac{1}{2}\times0.704\times32.065 \\ Mass\text{ of S}=11.29grams \end{gathered}[/tex]Hence the mass of magnesium oxide and the mass of sulphur that forms is 28.35grams and 11.29grams