Question

g Suppose 0.0350 g M g is reacted with 10.00 mL of 6 M H C l to produce aqueous magnesium chloride and hydrogen gas. M g ( s ) + 2 H C l ( a q ) → M g C l 2 ( a q ) + H 2 ( g ) What is the limiting reactant in this reaction?

Answer 1

Answer:

Mg will be the limiting reagent.

Explanation:

The balanced reaction is:

Mg + 2 HCl → MgCl₂ + H₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

• Mg: 1 mole
• HCl: 2 moles
• MgCl₂: 1 mole
• H₂: 1 mole

Being the molar mass of each compound:

• Mg: 24.3 g/mole
• HCl: 36.45 g/mole
• MgCl₂: 95.2 g/mole
• H₂: 2 g/mole

By reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

• Mg: 1 mole* 24.3 g/mole= 24.3 g
• HCl: 2 moles* 36.45 g/mole= 72.9 g
• MgCl₂: 1 mole* 95.2 g/mole= 95.2 g
• H₂: 1 mole* 2 g/mole= 2 g

0.0350 g of Mg is reacted with 10.00 mL (equal to 0.01 L) of 6 M HCl.

Molarity being the number of moles of solute that are dissolved in a certain volume, expressed as:

$Molarity=\frac{number of moles of solute}{volume}$

in units $\frac{moles}{liter}$

then, the number of moles of HCl that react is:

$6 M=\frac{number of moles of HCl}{0.01 L}$

number of moles of HCl= 6 M*0.01 L

number of moles of HCl= 0.06 moles

Then you can apply the following rule of three: if by stoichiometry 2 moles of HCl react with 24.3 grams of Mg, 0.06 moles of HCl react with how much mass of Mg?

$mass of Mg=\frac{0.06 moles of HCl* 24.3 grams of Mg}{2 moles of HCl}$

mass of Mg= 0.729 grams

But 0.729 grams of Mg are not available, 0.0350 grams are available. Since you have less mass than you need to react with 0.06 moles of HCl, Mg will be the limiting reagent.