
For the reaction represented by the equation $C{{l}_{2}}+2KBr\to 2KCl+B{{r}_{2}}$, how many grams of potassium chloride can be produced from 300 g each of chlorine and potassium bromide?
Answer
532.2k+ views
Hint: Since the equation is balanced, find out the number of moles of chlorine and potassium bromide and find out limiting reagent. With the help of stoichiometric coefficient, find out the mass of KCl.
Complete step by step answer:
In order to answer our question, we need to learn about stoichiometry and limiting reagent. We also need to learn about the molar concept and limiting reagent. Now, matter is made up of atoms and what differentiates atoms of two elements is its mass. Now, by generalisation, $6\times {{10}^{23}}$atoms of any element taken together is called 1 mole of the substance. But, the mass of 1 mole of an element will not match with the mass of 1 mole of another element. This mass is called the molar mass and the huge number is called the Avogadro’s constant. The number of moles of a particular substance is given by the formula:
\[no\,of\,moles=\dfrac{given\,mass}{molar\,mass}\]
In a chemical reaction, preferably balanced, if there are two or more reactants and one of the reactants is taken in limited quantity, so after that particular reagent gets used up, the reaction no longer proceeds, this reagent is called the limiting reagent.
Now, with the help of the theory, we will solve the question. Let us calculate the moles of $C{{l}_{2}}$ and KBr:
\[\begin{align}
& mole\,C{{l}_{2}}=\dfrac{300g}{35.45g\,mo{{l}^{-1}}}=8.463\,mol \\
& mole\,KBr=\dfrac{300g}{119.01g\,mo{{l}^{-1}}}=2.521\,mol \\
\end{align}\]
The molecules of chlorine are way more than moles of KBr, so we can say that potassium bromide is acting as the limiting reagent in the reaction. As stoichiometric coefficients of KCl and KBr are the same, the same amount of moles will be taken for them. Now, the mass of KCl will be:
\[2.521\,mol\times 74.55\,g\,mo{{l}^{-1}}=187.9g\]
So, we get 187.9 grams of potassium chloride can be produced from 300 g each of chlorine and potassium bromide
Note:
It is to be noted that the limiting reagent can also be found out by determining the number of moles from a balanced chemical equation, and then finding out the mole ratio from the information.
Complete step by step answer:
In order to answer our question, we need to learn about stoichiometry and limiting reagent. We also need to learn about the molar concept and limiting reagent. Now, matter is made up of atoms and what differentiates atoms of two elements is its mass. Now, by generalisation, $6\times {{10}^{23}}$atoms of any element taken together is called 1 mole of the substance. But, the mass of 1 mole of an element will not match with the mass of 1 mole of another element. This mass is called the molar mass and the huge number is called the Avogadro’s constant. The number of moles of a particular substance is given by the formula:
\[no\,of\,moles=\dfrac{given\,mass}{molar\,mass}\]
In a chemical reaction, preferably balanced, if there are two or more reactants and one of the reactants is taken in limited quantity, so after that particular reagent gets used up, the reaction no longer proceeds, this reagent is called the limiting reagent.
Now, with the help of the theory, we will solve the question. Let us calculate the moles of $C{{l}_{2}}$ and KBr:
\[\begin{align}
& mole\,C{{l}_{2}}=\dfrac{300g}{35.45g\,mo{{l}^{-1}}}=8.463\,mol \\
& mole\,KBr=\dfrac{300g}{119.01g\,mo{{l}^{-1}}}=2.521\,mol \\
\end{align}\]
The molecules of chlorine are way more than moles of KBr, so we can say that potassium bromide is acting as the limiting reagent in the reaction. As stoichiometric coefficients of KCl and KBr are the same, the same amount of moles will be taken for them. Now, the mass of KCl will be:
\[2.521\,mol\times 74.55\,g\,mo{{l}^{-1}}=187.9g\]
So, we get 187.9 grams of potassium chloride can be produced from 300 g each of chlorine and potassium bromide
Note:
It is to be noted that the limiting reagent can also be found out by determining the number of moles from a balanced chemical equation, and then finding out the mole ratio from the information.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Class 11 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Trending doubts
10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Explain zero factorial class 11 maths CBSE

What is a periderm How does periderm formation take class 11 biology CBSE

