If a reaction mixture contains 24 g of Mg and 32 g of O2 , what is the limiting reactant?

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This problem is providing the mass of both magnesium metal and oxygen gas and involved in a chemical reaction and asks for the limiting reactant. At the end, it turns out to be identified as magnesium.

<h3>Stoichiometry</h3>

In chemistry, stoichiometry is a widely-used tool we use in order to relate the mass and moles of different chemical substances involved in a chemical reaction. Thus, we consider the following chemical equation between magnesium and oxygen to produce magnesium oxide.

$2Mg+O_2\rightarrow 2MgO$

However, when the mass of the both of the reactants is given, one must identify the limiting reactant as the one producing the least of the moles of the product, which means we can use the given grams of the both of the reactants, their molar masses and mole ratios with the product to obtain the aforementioned:

$24gMg*\frac{1molMg}{24.3gMg}*\frac{2molMgO}{2molMg}=0.988molMgO\\ \\ 32gO_2*\frac{1molO_2}{32.0gO_2}*\frac{2molMgO}{1molO_2}=2molMgO$

Thus, we can evidence how 24 g of magnesium produce the least of the moles of magnesium oxide, fact validating the magnesium as the limiting reactant and the oxygen as the excess one.