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This is one of our many quizzes on Limiting reactants. Click on this to access more such competitive quizzes on Physical Chemistry.

Limiting reactants Quiz 2

chemistry

 When X reacted with Y then Z was produced but some amount of X was also there while no trace of Y was present. So, X is:

 If the reaction formed is less than the expected value, then one the reactant must be in lesser amount while the other reactant will be in excess amount. 

In the reaction, X and Y react to Z with some X was also present which means X is an excess reagent and some part of it is unreacted while the Y is the limiting reagent because it is fully consumed in the reaction.

 To calculate the limiting reactant, there are some steps to be followed, so what is the first step?

 If the number of moles in the given reaction will be wrong then we cannot predict the ratio in which the given mass of the reactants combine to form the products.

The steps followed are: write the equation or balance the given equation, then calculate the molecular mass, calculate the number of moles for the given mass, then find the limiting and excess reagent.

A reaction is given: $2A+3B\to C$ Calculate the amount of C when 20 moles of A and 25 moles of B react:

  First, find the limiting reagent with respect to the given number of moles and then take the ratio of limiting reagent and product formed.

The given reaction is: $\\ 2A+3B\to C\\ $ As we can see that 2 moles of A react with 3 moles of B, so when we take 20 moles of A then moles of B should be 30. But the value of B given is 25, therefore, B is a limiting reagent. Now, 3 moles of B react to form 1 mole of C, when 25 moles of B are taken then C will be: $\\ C=\dfrac{25}{3}=8.3\text{ moles}$

A given reaction is: $CaC{{O}_{3}}+2HCl\to CaC{{l}_{2}}+{{H}_{2}}O+C{{O}_{2}}$ When 5 grams of $CaC{{O}_{3}}$ and 5 grams of HCl reacts, then which one will be limiting reagent?

 Calculate the number of moles for the given mass of the compounds and then compare the ratio.

 The given reaction is: $\\ CaC{{O}_{3}}+2HCl\to CaC{{l}_{2}}+{{H}_{2}}O+C{{O}_{2}}\\ $ As we can see that in the reaction, 1 mole of calcium carbonate reacts with 2 moles of hydrochloric acid. The given mass is 5 grams of $CaC{{O}_{3}}$ and 5 grams of HCl will react. The molecular mass of $CaC{{O}_{3}}$ is 100, so the number of moles will be 0.05. The molecular mass of HCl is 36.5, so the number of moles will be 0.13. $ \\ $ So, for 0.13 moles of HCl, we require 0.65 moles of $CaC{{O}_{3}}$, but the given amount of $\\ CaC{{O}_{3}}$ is 0.05. Therefore, $CaC{{O}_{3}}$ is the limiting reagent.

<img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/ecaae2e5-7a66-4700-a1a1-4a8949176a481161029478782129222.png"/> $ \\ $ In the above reaction, 10 moles of ${{H}_{2}}$ were taken and 15 moles of $C{{l}_{2}}$ were taken. Find out which one of them is an excess reactant and why?

  First, find the excess reagent with respect to the given number of moles and then find the unreacted by calculating the moles required for the limiting reagent. 

 According to the diagram, equal moles of hydrogen and chlorine react to form hydrochloric acid. So, if the 10 moles of ${{H}_{2}}$ is taken then 10 moles of $C{{l}_{2}}$ will be required, but the given amount of $C{{l}_{2}}$ is 15 moles. Therefore, $C{{l}_{2}}$ is the excess reagent and amount left will be: $ \\ 15 – 10 = 5 \text{moles}$