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Question: For the reaction represented by the equation \(C{{l}_{2}}+2KBr\to 2KCl+B{{r}_{2}}\), how many grams ...

For the reaction represented by the equation Cl2+2KBr→2KCl+Br2C{{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?

Explanation

Solution

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, 6×10236\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=given massmolar massno\,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 Cl2C{{l}_{2}} and KBr:

& 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{aligned}$$ The molecules of chlorine are way more than moles of KBr, so we can say that potassium bromide is acting as the limiting reagent in the reaction. As stoichiometric coefficients of KCl and KBr are the same, the same amount of moles will be taken for them. Now, the mass of KCl will be: $$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.