iGCSE Chemistry (0620) Core:3.3 The mole and the Avogadro constant: Exam Style Questions Paper 1- New Syllabus

To compute the reacting mass, we first evaluate the moles of reacting Magnesium using its relative atomic mass ($A_r(\text{Mg}) = 24$). Dividing the given mass by its atomic mass ($\text{moles} = 48\text{ g} / 24\text{ g/mol}$) yields exactly $2\text{ moles}$ of $\text{Mg}$. According to the balanced stoichiometric equation, the molar ratio between $\text{Mg}$ and $\text{MgO}$ is a direct $2:2$ (or $1:1$) relationship, meaning $2\text{ moles}$ of $\text{Mg}$ will yield exactly $2\text{ moles}$ of $\text{MgO}$. The relative formula mass of magnesium oxide is calculated as $M_r(\text{MgO}) = 24 + 16 = 40$. Multiplying the number of moles by this formula mass ($\text{mass} = 2\text{ moles} \times 40\text{ g/mol}$) results in $80\text{ g}$ of $\text{MgO}$, making option C the correct answer.
✅ Answer: (C)