Chemistry, 04.03.2020 23:43 angel213326

Magnesium (used in the manufacture of light alloys) reacts with iron(III) chloride to form magnesium chloride and iron. A mixture of 41.0 g of magnesium and 175.0 g of iron(III) chloride is allowed to react. Identify the limiting reactant and determine the mass of the excess reactant present in the vessel when the reaction is complete.

Answers: 3

Show answers

Answers

Answer from: robert7248

The limiting reactant is magnesium and mass of excess reactant present in the vessel is 96.35 grams.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For magnesium:

Given mass of magnesium = 41.0 g

Molar mass of magnesium = 24 g/mol

Putting values in equation 1, we get:

$\text{Moles of magnesium}=\frac{41.0g}{24g/mol}=1.708mol$

For iron(III) chloride:

Given mass of iron(III) chloride = 175.0 g

Molar mass of iron(III) chloride = 162.2 g/mol

Putting values in equation 1, we get:

$\text{Moles of iron(III) chloride}=\frac{175g}{162.2g/mol}=1.708mol$

The chemical equation for the reaction of magnesium and iron(III) chloride follows:

$3Mg+2FeCl_3\rightarrow 3MgCl_2+2Fe$

By Stoichiometry of the reaction:

3 moles of magnesium reacts with 2 moles of iron(III) chloride

So, 1.708 moles of magnesium will react with = $\frac{2}{3}\times 1.708=1.114mol$ of iron(III) chloride

As, given amount of iron(III) chloride is more than the required amount. So, it is considered as an excess reagent.

Thus, magnesium is considered as a limiting reagent because it limits the formation of product.

Moles of excess reactant left (iron(III) chloride) = [1.708 - 1.114] = 0.594 moles

Now, calculating the mass of iron(III) chloride from equation 1, we get:

Molar mass of iron(III) chloride = 162.2 g/mol

Moles of iron(III) chloride = 0.594 moles

Putting values in equation 1, we get:

$0.594mol=\frac{\text{Mass of iron(III) chloride}}{162.2g/mol}\\\\\text{Mass of iron(III) chloride}=(0.594mol\times 162.2g/mol)=96.35g$

Hence, the limiting reactant is magnesium and mass of excess reactant present in the vessel is 96.35 grams.

Answer from: Quest

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.

Answer from: Quest

What’s the solution?

Another question on Chemistry

Chemistry, 22.06.2019 03:00

Which of the dna typing techniques do you think you would choose if you had to analyze a dna sample? why?

Answers: 1

Answer

Chemistry, 22.06.2019 08:20

What is the formula for the compound dinitrogen pentoxide? a. n4o5 b. n5o4 c. n4o6 d. n5o2 e. n2o5

Answers: 3

Answer

Chemistry, 22.06.2019 13:30

An animal cell loses the ability to convert energy stored in food to energy that the cell can use. which of the cell's organelles has stopped working? a.the mitochondria b.the nucleus c.the vacuoles d.the endoplasmic reticulum

Answers: 1

Answer

Chemistry, 22.06.2019 13:50

What happens when an atom of sulfur combines with two atoms of chlorine to produce sci2? a. each chlorine atom shares a pair of electrons with the sulfur atom. b. an electron is transferred from each chlorine atom to the sulfur atom. c. an electron is transferred from the sulfur atom to each chlorine atom. d. each chlorine atom shares all its valence electrons with the sulfur atom.

Answers: 2

Answer

You know the right answer?

Magnesium (used in the manufacture of light alloys) reacts with iron(III) chloride to form magnesium...

Questions