
Calculate the number of millimoles of ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ which will completely react with 40 ml 0.1 M KI solution.
Answer
558.6k+ views
Hint: The solution for the given problem depends on the basic knowledge of stoichiometry and molarity. Knowing about these topics and properly balancing the chemical equation, we can solve the given illustration.
Complete Solution :
Let us know about some of the basic terms before solving the given illustration;
Balancing the chemical equation-
This is the primary step before analysing any chemical reaction and concluding something out of it.
- Stoichiometric coefficients-
These are the coefficients used to balance the equation. We change the stoichiometric coefficients while we balance the equation, not the subscripts of the compounds involved.
- Molarity-
Molarity is defined as the moles of solute per litre of solution. In short, it is the amount of the substance in a certain volume of solution. It is stated as,
\[M=\dfrac{n}{V}\]
where,
M = molar concentration
n = moles of solute
V = litres of solution
Unit of molarity is mol/lit i.e. M.
Thus, now let us move towards the illustration,
Given data:
Volume of KI = 40 ml
Molarity of KI solution = 0.1 M
The reaction can be stated as,
${{K}_{2}}C{{r}_{2}}{{O}_{7}}+7{{H}_{2}}S{{O}_{4}}+6KI\to C{{r}_{2}}{{\left( S{{O}_{4}} \right)}_{3}}+3{{I}_{2}}+4{{K}_{2}}S{{O}_{4}}+7{{H}_{2}}O$
- Now, according to the stoichiometry,
1 mole of ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ $\equiv $ 6 moles of KI
And, in accordance with given data we can state that,
x moles of ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$$\equiv $ $\dfrac{40}{1000}\times 0.1$ moles of KI
- Thus, by solving the above equations we can state that,
$x=\dfrac{40\times 0.1}{1000\times 6}$ moles of ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$
x = 0.00066 moles of ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$
Therefore, we can say that 0.66 millimoles of ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ will completely react with 40 ml 0.1 M KI solution.
Note: Do note to use proper units while solving the illustrations as above. Any mistake while calculating will result in chaos which is undesirable.
Complete Solution :
Let us know about some of the basic terms before solving the given illustration;
Balancing the chemical equation-
This is the primary step before analysing any chemical reaction and concluding something out of it.
- Stoichiometric coefficients-
These are the coefficients used to balance the equation. We change the stoichiometric coefficients while we balance the equation, not the subscripts of the compounds involved.
- Molarity-
Molarity is defined as the moles of solute per litre of solution. In short, it is the amount of the substance in a certain volume of solution. It is stated as,
\[M=\dfrac{n}{V}\]
where,
M = molar concentration
n = moles of solute
V = litres of solution
Unit of molarity is mol/lit i.e. M.
Thus, now let us move towards the illustration,
Given data:
Volume of KI = 40 ml
Molarity of KI solution = 0.1 M
The reaction can be stated as,
${{K}_{2}}C{{r}_{2}}{{O}_{7}}+7{{H}_{2}}S{{O}_{4}}+6KI\to C{{r}_{2}}{{\left( S{{O}_{4}} \right)}_{3}}+3{{I}_{2}}+4{{K}_{2}}S{{O}_{4}}+7{{H}_{2}}O$
- Now, according to the stoichiometry,
1 mole of ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ $\equiv $ 6 moles of KI
And, in accordance with given data we can state that,
x moles of ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$$\equiv $ $\dfrac{40}{1000}\times 0.1$ moles of KI
- Thus, by solving the above equations we can state that,
$x=\dfrac{40\times 0.1}{1000\times 6}$ moles of ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$
x = 0.00066 moles of ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$
Therefore, we can say that 0.66 millimoles of ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ will completely react with 40 ml 0.1 M KI solution.
Note: Do note to use proper units while solving the illustrations as above. Any mistake while calculating will result in chaos which is undesirable.
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