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

3.98 g

Step-by-step 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) = (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) = (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) = (0.0987~mol)/(1) = 0.0987~mol`

Equivalent of oxygen:

`eq_(O_2) = (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`