
A particular \[100\] octane aviation gasoline used \[{\rm{1}}{\rm{.00}}\,\,{\rm{mL}}\] of tetraethyllead, \[{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}}\]of density \[{\rm{1}}{\rm{.66}}\,\,\dfrac{{\rm{g}}}{{{\rm{mL}}}}\]per litre of product. This compound is made as follows:
\[{\rm{4}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}}\,\,{\rm{ + }}\,\,{\rm{4NaPb}} \to {{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}}\,\,{\rm{ + }}\,\,{\rm{4NaCl}}\,\,{\rm{ + }}\,\,{\rm{3Pb}}\]
How many grams of ethyl chloride is needed to make enough tetraethyl lead for \[{\rm{1}}\,\,{\rm{L}}\] of gasoline? [Atomic mass of \[{\rm{Pb}}\,\,{\rm{ = }}\,\,{\rm{207}}\]]
Answer
162.6k+ views
Hint: Density of tetraethyllead is required to calculate first followed by the number of moles tetraethyllead. Number of moles of tetraethyllead is calculated by using its molar mass. The molar ratio from the balanced molecular equation is used to find the moles of ethyl chloride and then its mass in grams is calculated with the help of its molar mass.
Complete Step by Step Solution:
Density of a substance is defined as the mass per unit volume.
Mathematically, \[{\rm{density}}\,{\rm{ = }}\,\dfrac{{{\rm{mass}}}}{{{\rm{volume}}}}\]
As per given data,
Volume of tetraethyllead, \[{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}}\] \[{\rm{ = }}\,{\rm{1}}{\rm{.00}}\,\,{\rm{mL}}\]
Density of tetraethyllead, \[{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}}\] \[{\rm{ = }}\,\,{\rm{1}}{\rm{.66}}\,\,\dfrac{{\rm{g}}}{{{\rm{mL}}}}\]
Find the mass of tetraethyllead, \[{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}}\]as shown below:
\[\begin{array}{l}{\rm{density}}\,{\rm{ = }}\,\dfrac{{{\rm{mass}}}}{{{\rm{volume}}}}\\ \Rightarrow {\rm{mass}}\,\,{\rm{of}}\,\,{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}} = \,\,{\rm{density}}\,\,{\rm{of}}\,\,{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}}\,\,{\rm{ \times }}\,\,{\rm{volume}}\,\,{\rm{of}}\,\,{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}}\\ \Rightarrow {\rm{mass}}\,\,{\rm{of}}\,\,{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}} = \,\,{\rm{1}}{\rm{.66}}\,\,\dfrac{{\rm{g}}}{{{\rm{mL}}}} \times {\rm{1}}{\rm{.00}}\,\,{\rm{mL}}\\ \Rightarrow {\rm{mass}}\,\,{\rm{of}}\,\,{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}} = \,\,{\rm{1}}{\rm{.66}}\,\,{\rm{g}}\end{array}\]
Molar mass of tetraethyllead, \[{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}}\]is known to be \[{\rm{323}}{\rm{.4}}\dfrac{{\rm{g}}}{{{\rm{mol}}}}\]
The relationship between moles (n), mass (m), and molar mass (M) is as shown below:
\[{\rm{moles}}\,{\rm{(n) = }}\dfrac{{{\rm{mass(m)}}}}{{{\rm{molar}}\,\,{\rm{mass}}\,\,{\rm{(M)}}}}\]
Find the moles of tetraethyllead, \[{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}}\] as shown below:
\[\begin{array}{l}{\rm{moles}}\,{\rm{(n)}}\,\,{\rm{of}}\,\,{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}} = \dfrac{{{\rm{mass}}\,\,{\rm{(m)}}\,\,{\rm{of}}\,\,{{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}}_{\rm{4}}}{\rm{Pb}}}}{{{\rm{molar}}\,\,{\rm{mass}}\,\,{\rm{(M)}}\,\,{\rm{of}}\,\,{{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}}_{\rm{4}}}{\rm{Pb}}}}\\ \Rightarrow {\rm{moles}}\,{\rm{(n)}}\,\,{\rm{of}}\,\,{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}} = \dfrac{{{\rm{1}}{\rm{.66}}\,\,{\rm{g}}}}{{{\rm{323}}{\rm{.4}}\dfrac{{\rm{g}}}{{{\rm{mol}}}}}}\\ \Rightarrow {\rm{moles}}\,{\rm{(n)}}\,\,{\rm{of}}\,\,{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}} = \,\,{\rm{0}}{\rm{.0051}}\,\,{\rm{mol}}\end{array}\]
The balanced molecular equation is given as:
\[{\rm{4}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}}\,\,{\rm{ + }}\,\,{\rm{4NaPb}} \to {{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}}\,\,{\rm{ + }}\,\,{\rm{4NaCl}}\,\,{\rm{ + }}\,\,{\rm{3Pb}}\]
From the above equation, it can be seen that \[4\]moles of ethyl chloride, \[{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}}\]reacts to form \[{\rm{1}}\]moles of tetraethyllead, \[{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}}\].
So, molar ratio of \[{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}}\,:{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}} = 4\,\,:\,\,1\]
Find the moles of ethyl chloride, \[{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}}\] from the calculated moles of tetraethyllead, \[{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}}\] and their molar ratio as:
\[{\rm{0}}{\rm{.0051}}\,\,{\rm{mol}}\,\,\,{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb \times }}\dfrac{{{\rm{4}}\,\,{\rm{mol}}\,\,{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}}}}{{{\rm{1}}\,\,{\rm{mol}}\,\,{{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}}_{\rm{4}}}{\rm{Pb}}}}\\ = \,0.0204\,\,{\rm{mol}}\,\,{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}}\]
Molar mass of ethyl chloride, \[{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}}\] is known to be \[{\rm{64}}{\rm{.51}}\dfrac{{\rm{g}}}{{{\rm{mol}}}}\]
Lastly, find the mass of ethyl chloride, \[{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}}\] from its calculated moles by using the molar mass as:
\[\begin{array}{l}{\rm{moles}}\,{\rm{(n)}}\,\,{\rm{of}}\,\,{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}}\, = \dfrac{{{\rm{mass}}\,\,{\rm{(m)}}\,\,{\rm{of}}\,\,{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}}\,}}{{{\rm{molar}}\,\,{\rm{mass}}\,\,{\rm{(M)}}\,\,{\rm{of}}\,\,{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}}\,}}\\ \Rightarrow {\rm{mass}}\,\,{\rm{(m)}}\,\,{\rm{of}}\,\,{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}} = {\rm{moles}}\,{\rm{(n)}}\,\,{\rm{of}}\,\,{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}} \times {\rm{molar}}\,\,{\rm{mass}}\,\,{\rm{(M)}}\,\,{\rm{of}}\,\,{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}}\\ \Rightarrow {\rm{mass}}\,\,{\rm{(m)}}\,\,{\rm{of}}\,\,{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}} = \,0.0204\,\,{\rm{mol}} \times {\rm{64}}{\rm{.51}}\dfrac{{\rm{g}}}{{{\rm{mol}}}}\\ \Rightarrow {\rm{mass}}\,\,{\rm{(m)}}\,\,{\rm{of}}\,\,{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}} = \,1.32\,\,{\rm{g}}\end{array}\]
Hence, the mass of ethyl chloride, \[{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}}\] is found to be \[1.32\,\,{\rm{g}}\]
Therefore, the answer is \[\,1.32\,\,{\rm{g}}\,\,{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}}\]
Note: The calculation of relative amounts of substances in chemical reactions is known as stoichiometry (derived from the Greek stoicheion meaning element and meterin meaning measures). The key to weight relationship in chemical change is the mole concept. Coefficients placed in front of formulae in a balanced chemical equation represent the ratio by moles in which substances are consumed and produced by the chemical change. Since the mass of a mole of a substance is directly related to its formula or molecular weight, a balanced chemical equation represents a ratio by weight also.
Complete Step by Step Solution:
Density of a substance is defined as the mass per unit volume.
Mathematically, \[{\rm{density}}\,{\rm{ = }}\,\dfrac{{{\rm{mass}}}}{{{\rm{volume}}}}\]
As per given data,
Volume of tetraethyllead, \[{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}}\] \[{\rm{ = }}\,{\rm{1}}{\rm{.00}}\,\,{\rm{mL}}\]
Density of tetraethyllead, \[{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}}\] \[{\rm{ = }}\,\,{\rm{1}}{\rm{.66}}\,\,\dfrac{{\rm{g}}}{{{\rm{mL}}}}\]
Find the mass of tetraethyllead, \[{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}}\]as shown below:
\[\begin{array}{l}{\rm{density}}\,{\rm{ = }}\,\dfrac{{{\rm{mass}}}}{{{\rm{volume}}}}\\ \Rightarrow {\rm{mass}}\,\,{\rm{of}}\,\,{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}} = \,\,{\rm{density}}\,\,{\rm{of}}\,\,{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}}\,\,{\rm{ \times }}\,\,{\rm{volume}}\,\,{\rm{of}}\,\,{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}}\\ \Rightarrow {\rm{mass}}\,\,{\rm{of}}\,\,{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}} = \,\,{\rm{1}}{\rm{.66}}\,\,\dfrac{{\rm{g}}}{{{\rm{mL}}}} \times {\rm{1}}{\rm{.00}}\,\,{\rm{mL}}\\ \Rightarrow {\rm{mass}}\,\,{\rm{of}}\,\,{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}} = \,\,{\rm{1}}{\rm{.66}}\,\,{\rm{g}}\end{array}\]
Molar mass of tetraethyllead, \[{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}}\]is known to be \[{\rm{323}}{\rm{.4}}\dfrac{{\rm{g}}}{{{\rm{mol}}}}\]
The relationship between moles (n), mass (m), and molar mass (M) is as shown below:
\[{\rm{moles}}\,{\rm{(n) = }}\dfrac{{{\rm{mass(m)}}}}{{{\rm{molar}}\,\,{\rm{mass}}\,\,{\rm{(M)}}}}\]
Find the moles of tetraethyllead, \[{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}}\] as shown below:
\[\begin{array}{l}{\rm{moles}}\,{\rm{(n)}}\,\,{\rm{of}}\,\,{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}} = \dfrac{{{\rm{mass}}\,\,{\rm{(m)}}\,\,{\rm{of}}\,\,{{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}}_{\rm{4}}}{\rm{Pb}}}}{{{\rm{molar}}\,\,{\rm{mass}}\,\,{\rm{(M)}}\,\,{\rm{of}}\,\,{{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}}_{\rm{4}}}{\rm{Pb}}}}\\ \Rightarrow {\rm{moles}}\,{\rm{(n)}}\,\,{\rm{of}}\,\,{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}} = \dfrac{{{\rm{1}}{\rm{.66}}\,\,{\rm{g}}}}{{{\rm{323}}{\rm{.4}}\dfrac{{\rm{g}}}{{{\rm{mol}}}}}}\\ \Rightarrow {\rm{moles}}\,{\rm{(n)}}\,\,{\rm{of}}\,\,{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}} = \,\,{\rm{0}}{\rm{.0051}}\,\,{\rm{mol}}\end{array}\]
The balanced molecular equation is given as:
\[{\rm{4}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}}\,\,{\rm{ + }}\,\,{\rm{4NaPb}} \to {{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}}\,\,{\rm{ + }}\,\,{\rm{4NaCl}}\,\,{\rm{ + }}\,\,{\rm{3Pb}}\]
From the above equation, it can be seen that \[4\]moles of ethyl chloride, \[{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}}\]reacts to form \[{\rm{1}}\]moles of tetraethyllead, \[{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}}\].
So, molar ratio of \[{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}}\,:{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}} = 4\,\,:\,\,1\]
Find the moles of ethyl chloride, \[{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}}\] from the calculated moles of tetraethyllead, \[{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb}}\] and their molar ratio as:
\[{\rm{0}}{\rm{.0051}}\,\,{\rm{mol}}\,\,\,{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}_{\rm{4}}}{\rm{Pb \times }}\dfrac{{{\rm{4}}\,\,{\rm{mol}}\,\,{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}}}}{{{\rm{1}}\,\,{\rm{mol}}\,\,{{{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}}_{\rm{4}}}{\rm{Pb}}}}\\ = \,0.0204\,\,{\rm{mol}}\,\,{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}}\]
Molar mass of ethyl chloride, \[{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}}\] is known to be \[{\rm{64}}{\rm{.51}}\dfrac{{\rm{g}}}{{{\rm{mol}}}}\]
Lastly, find the mass of ethyl chloride, \[{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}}\] from its calculated moles by using the molar mass as:
\[\begin{array}{l}{\rm{moles}}\,{\rm{(n)}}\,\,{\rm{of}}\,\,{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}}\, = \dfrac{{{\rm{mass}}\,\,{\rm{(m)}}\,\,{\rm{of}}\,\,{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}}\,}}{{{\rm{molar}}\,\,{\rm{mass}}\,\,{\rm{(M)}}\,\,{\rm{of}}\,\,{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}}\,}}\\ \Rightarrow {\rm{mass}}\,\,{\rm{(m)}}\,\,{\rm{of}}\,\,{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}} = {\rm{moles}}\,{\rm{(n)}}\,\,{\rm{of}}\,\,{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}} \times {\rm{molar}}\,\,{\rm{mass}}\,\,{\rm{(M)}}\,\,{\rm{of}}\,\,{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}}\\ \Rightarrow {\rm{mass}}\,\,{\rm{(m)}}\,\,{\rm{of}}\,\,{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}} = \,0.0204\,\,{\rm{mol}} \times {\rm{64}}{\rm{.51}}\dfrac{{\rm{g}}}{{{\rm{mol}}}}\\ \Rightarrow {\rm{mass}}\,\,{\rm{(m)}}\,\,{\rm{of}}\,\,{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}} = \,1.32\,\,{\rm{g}}\end{array}\]
Hence, the mass of ethyl chloride, \[{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}}\] is found to be \[1.32\,\,{\rm{g}}\]
Therefore, the answer is \[\,1.32\,\,{\rm{g}}\,\,{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{Cl}}\]
Note: The calculation of relative amounts of substances in chemical reactions is known as stoichiometry (derived from the Greek stoicheion meaning element and meterin meaning measures). The key to weight relationship in chemical change is the mole concept. Coefficients placed in front of formulae in a balanced chemical equation represent the ratio by moles in which substances are consumed and produced by the chemical change. Since the mass of a mole of a substance is directly related to its formula or molecular weight, a balanced chemical equation represents a ratio by weight also.
Recently Updated Pages
Two pi and half sigma bonds are present in A N2 + B class 11 chemistry JEE_Main

Which of the following is most stable A Sn2+ B Ge2+ class 11 chemistry JEE_Main

The enolic form of acetone contains a 10sigma bonds class 11 chemistry JEE_Main

The specific heat of metal is 067 Jg Its equivalent class 11 chemistry JEE_Main

The increasing order of a specific charge to mass ratio class 11 chemistry JEE_Main

Which one of the following is used for making shoe class 11 chemistry JEE_Main

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

JoSAA JEE Main & Advanced 2025 Counselling: Registration Dates, Documents, Fees, Seat Allotment & Cut‑offs

NIT Cutoff Percentile for 2025

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Chemistry In Hindi Chapter 1 Some Basic Concepts of Chemistry

NCERT Solutions for Class 11 Chemistry Chapter 7 Redox Reaction

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

Degree of Dissociation and Its Formula With Solved Example for JEE
