Question

17.1 grams of magnesium metal burns in sulphur dioxide to form magnesium oxide and sulphur write a balanced equation for the reaction and calculate the mass of magnesium oxide and the mass of sulphur that forms.

Answer 1

Mass of MgO = 28.35grams

Mass of Sulphur = 11.29 grams

Explanations:

The balanced chemical equation between magnesium metal and sulphur dioxide is given as:

`2Mg+SO_2\rightarrow2MgO+S`

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:

`\begin{gathered} Mass\text{ of MgO}=0.704*40.3 \\ Mass\text{ of MgO}=28.35grams \end{gathered}`

Similarly, 2moles of Mg produces 1 mole of sulphur, hence the mass of sulphur produced is;

`\begin{gathered} Mass\text{ of S}=(1)/(2)*0.704*32.065 \\ Mass\text{ of S}=11.29grams \end{gathered}`

Hence the mass of magnesium oxide and the mass of sulphur that forms is 28.35grams and 11.29grams