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?

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:

[tex]Molarity=\frac{number of moles of solute}{volume}[/tex]

in units [tex]\frac{moles}{liter}[/tex]

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

[tex]6 M=\frac{number of moles of HCl}{0.01 L}[/tex]

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?

[tex]mass of Mg=\frac{0.06 moles of HCl* 24.3 grams of Mg}{2 moles of HCl}[/tex]

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.

Abdulazeez10 Abdulazeez10 · Answer 2 · 2021-11-19T16:19:17+01:00

The limiting reactant in the reaction is Magnesium (Mg)

From the question,

We are to determine the limiting reactant in the reaction.

The given balanced chemical equation for the reaction is

Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

This means

1 mole of Mg is required to react completely with 2 moles of HCl

Now, we will determine the number of moles of each reactant present

For Magnesium (Mg)

Mass = 0.0350 g

Using the formula

[tex]Number\ of\ moles = \frac{Mass}{Atomic\ mass}[/tex]

Atomic mass of Mg = 24.305 g/mol

∴ Number of moles of Mg present = [tex]\frac{0.0350}{24.305}[/tex]

Number of moles of Mg present = 0.00144 mole

For HCl

Concentration = 6M

Volume = 10.00 mL = 0.01 L

Using the formula

Number of moles = Concentration × Volume

∴ Number of moles HCl present = 6 × 0.01

Number of moles HCl present = 0.06 mole

Since,

1 mole of Mg is required to react completely with 2 moles of HCl

Then

0.00144 mole of Mg is required to react completely with 2×0.00144 mole of HCl

2×0.00144 = 0.00288

∴ The number of moles of HCl required to react completely with the Mg is 0.00288 mole

Since the number of moles of HCl present is more than 0.00288 mole, then HCl is the excess reactant and Mg is the limiting reactant.

Hence, the limiting reactant in the reaction is Magnesium (Mg)

Learn more here: https://brainly.com/question/13979150