Question

20ml of 0.1M $BaC{{l}_{2}}$ is mixed with 30 ml of 0.2M $A{{l}_{2}}{{(S{{O}_{4}})}_{3}}$. How many moles of $BaS{{O}_{4}}$ precipitated?

Answer 1

To solve this type of question we must have prior knowledge about different ways to calculate the number of moles. one way to calculate the number of moles is by finding the ratio of given mass to the molar mass.

Complete Solution :
Given in the question,
Molarity of $BaC{{l}_{2}}$= 0.1 M
Volume of $BaC{{l}_{2}}$= 20 ml
Molarity of $A{{l}_{2}}{{(S{{O}_{4}})}_{3}}$ = 0.2M
Volume of $A{{l}_{2}}{{(S{{O}_{4}})}_{3}}$ = 30 ml
- Moles can be calculated by the product of molarity and volume of the compound.
Moles of $BaC{{l}_{2}}$ = $0.1\times 2.0 = 2$ millimoles
Moles of $A{{l}_{2}}{{(S{{O}_{4}})}_{3}}$ = $0.2\times 3.0 = 6$ millimoles
The chemical reaction which will be involved in the process is mentioned below: $3BaC{{l}_{2}}+A{{l}_{2}}{{(S{{O}_{4}})}_{3}}\to 3BaS{{O}_{4}}+2AlC{{l}_{3}}$

- Using this equation, we can see that 3 moles of barium chloride is required to react with 2 moles of aluminium sulphate to produce 3 moles of barium sulphate and 2 moles of aluminium chloride.
But here 2 moles of barium chloride is present which means that barium chloride is the limiting reagent since it is present in less amount.
- So, the number of moles of $BaS{{O}_{4}}$ formed will be = 2 millimoles or 0.002 mol.
Hence, if 20ml of 0.1M $BaC{{l}_{2}}$ is mixed with 30 ml of 0.2M $A{{l}_{2}}{{(S{{O}_{4}})}_{3}}$. 0.002 moles of $BaS{{O}_{4}}$ will be precipitated.

Note: While doing such types of questions which involve stoichiometric calculations it is most important to balance the reaction. If the chemical reaction is not balanced then the answer can never be accurate. Millimole is the amount of substance equal to a thousand of a mole also known as mmol.