Answer:
3.98 g
Explanation:
Step 1. Write the balanced chemical reaction. In this case, magnesium reacts with oxygen to produce magnesium oxide:
[tex]2~Mg(s) + O_2 (g)\rightarrow 2~MgO (s)[/tex]
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Step 2. Calculate the number of moles of magnesium:
[tex]n_{Mg} = \frac{2.4~g}{24.305~g/mol} = 0.0987~mol[/tex]
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Step 3. Calculate the number of moles of oxygen:
[tex]n_{O_2} = \frac{10.0~g}{32.00~g/mol} = 0.3125~mol[/tex]
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Step 4. Identify the limiting reactant comparing the equivalents. Equivalent of Mg:
[tex]eq_{Mg} = \frac{0.0987~mol}{1} = 0.0987~mol[/tex]
Equivalent of oxygen:
[tex]eq_{O_2} = \frac{0.3125~mol}{2} = 0.15625~mol[/tex]
Therefore, Mg is the limiting reactant.
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Step 5. According to the stoichiometry of this reaction:
[tex]n_{Mg} = n_{MgO} = 0.0987~mol[/tex]
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Step 6. Convert the number of moles of MgO into mass:
[tex]m_{MgO} = 0.0987~mol\cdot 40.304~g/mol = 3.98~g[/tex]
