Question

Suppose 2.4 g of Mg reacts with 10.0 g of O2, to create magnesium oxide. How much magnesium oxide is produced?

Answer 1

Answer:

3.98 g

Explanation:

Step 1. Write the balanced chemical reaction. In this case, magnesium reacts with oxygen to produce magnesium oxide:

$2~Mg(s) + O_2 (g)\rightarrow 2~MgO (s)$

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Step 2. Calculate the number of moles of magnesium:

$n_{Mg} = \frac{2.4~g}{24.305~g/mol} = 0.0987~mol$

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Step 3. Calculate the number of moles of oxygen:

$n_{O_2} = \frac{10.0~g}{32.00~g/mol} = 0.3125~mol$

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Step 4. Identify the limiting reactant comparing the equivalents. Equivalent of Mg:

$eq_{Mg} = \frac{0.0987~mol}{1} = 0.0987~mol$

Equivalent of oxygen:

$eq_{O_2} = \frac{0.3125~mol}{2} = 0.15625~mol$

Therefore, Mg is the limiting reactant.

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Step 5. According to the stoichiometry of this reaction:

$n_{Mg} = n_{MgO} = 0.0987~mol$

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Step 6. Convert the number of moles of MgO into mass:

$m_{MgO} = 0.0987~mol\cdot 40.304~g/mol = 3.98~g$