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Identify the limiting reagent when 6.00 moles HCl combines with 5.00 moles Mg to form MgCl2?

1) Mg
2) HCl
3) MgCl2
4) H₂

User Nikolay K
by
8.2k points

1 Answer

6 votes

Final answer:

The limiting reagent is HCl because it is consumed completely in the reaction before the Mg is completely consumed.

Step-by-step explanation:

The limiting reagent can be identified by comparing the stoichiometric ratios of the reactants involved. In the given reaction, the balanced equation is:

2 HCl + Mg ⟶ MgCl2 + H2

According to the equation, 2 moles of HCl react with 1 mole of Mg to produce 1 mole of MgCl2. To determine the limiting reagent, we need to compare the number of moles of each reactant.

Given:

6.00 moles HCl

5.00 moles Mg

From the stoichiometry of the reaction, 2 moles of HCl is required to react with 1 mole of Mg. Therefore, 1 mole of Mg requires 2 moles of HCl.

So, for 5.00 moles of Mg, the required moles of HCl would be:

(5.00 moles Mg) x (2 moles HCl/1 mole Mg) = 10.00 moles HCl

Since we only have 6.00 moles of HCl available, it is the limiting reagent.

User Dlb
by
8.5k points
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