Sure, let's solve the problem step-by-step:
1. Given Data:
- Mass of Magnesium (Mg): 10 grams
- Molar mass of Magnesium (Mg): 24.305 g/mol
- Molar mass of Magnesium Oxide (MgO): 40.304 g/mol
2. Calculate the moles of Magnesium (Mg):
- Moles of Mg = [tex]\(\frac{\text{mass of Mg}}{\text{molar mass of Mg}}\)[/tex]
- Moles of Mg = [tex]\(\frac{10 \text{ grams}}{24.305 \text{ g/mol}}\)[/tex]
- Moles of Mg ≈ 0.41143797572515944 moles
3. Reaction Stoichiometry:
- The balanced chemical equation for the reaction is:
[tex]\[
2 \text{Mg} (s) + \text{O}_2 (g) \rightarrow 2 \text{MgO} (s)
\][/tex]
- According to the balanced equation, 1 mole of Mg produces 1 mole of MgO.
- Therefore, Moles of MgO produced = Moles of Mg
- Moles of MgO produced ≈ 0.41143797572515944 moles
4. Calculate the mass of Magnesium Oxide (MgO) produced:
- Mass of MgO = Moles of MgO [tex]\(\times\)[/tex] Molar mass of MgO
- Mass of MgO = 0.41143797572515944 moles [tex]\(\times\)[/tex] 40.304 g/mol
- Mass of MgO ≈ 16.582596173626825 grams
So, the mass of MgO obtained by burning 10 grams of Mg in air is approximately 16.6 grams. Therefore, the correct answer is:
a. 16.6 gm
