Question

If x g of \[{\text{Mn}}{{\text{O}}_{\text{2}}}\] and the volume of \[{\text{HCl}}\] of specific gravity \[{\text{1}}{\text{.2 g m}}{{\text{l}}^{{\text{ - 3}}}}\] and 5% by weight needed to produce 1.12 L of \[{\text{C}}{{\text{l}}_{\text{2}}}\] at STP by the reaction,
\[{\text{Mn}}{{\text{O}}_{\text{2}}}{\text{ + 4HCl }} \to {\text{ MnC}}{{\text{l}}_{\text{2}}}{\text{ + 3}}{{\text{H}}_{\text{2}}}{\text{O + C}}{{\text{l}}_{\text{2}}}\]
Then value of 100x is :
A) 435
B) 450
C) 400
D) None of these

Answer 1

Hint: The condition STP indicates standard temperature (273K) and standard pressure (1 atm). At STP condition volume of 1 mole of gas is always 22.4L. Using the standard volume of \[{\text{C}}{{\text{l}}_{\text{2}}}\] at STP calculate the moles of\[{\text{C}}{{\text{l}}_{\text{2}}}\] gas having a volume of 1.12L. Using the calculated moles of \[{\text{C}}{{\text{l}}_{\text{2}}}\] and stoichiometric ratio of \[{\text{Mn}}{{\text{O}}_{\text{2}}}\] and \[{\text{C}}{{\text{l}}_{\text{2}}}\] calculate the moles of \[{\text{Mn}}{{\text{O}}_{\text{2}}}\]. Convert these moles of \[{\text{Mn}}{{\text{O}}_{\text{2}}}\] to mass of it using its molar mass. Finally calculate the value of 100x.

Complete step-by-step answer:
 The reaction between \[{\text{Mn}}{{\text{O}}_{\text{2}}}\]and \[{\text{HCl}}\] given to us is:
\[{\text{Mn}}{{\text{O}}_{\text{2}}}{\text{ + 4HCl }} \to {\text{ MnC}}{{\text{l}}_{\text{2}}}{\text{ + 3}}{{\text{H}}_{\text{2}}}{\text{O + C}}{{\text{l}}_{\text{2}}}\]
We have given 1.12 L of \[{\text{C}}{{\text{l}}_{\text{2}}}\]gas produced at STP when x g of \[{\text{Mn}}{{\text{O}}_{\text{2}}}\] reacts.
We know that at STP condition volume of 1 mol of any gas is 22.4L.
Now using this reaction we will calculate the moles of \[{\text{C}}{{\text{l}}_{\text{2}}}\]gas present in 1.12L.
\[{\text{Moles of C}}{{\text{l}}_{\text{2}}} = \dfrac{{{\text{1}}{\text{.12L }} \times {\text{ 1 mol }}}}{{22.4{\text{ L }}}} = 0.05{\text{ mol C}}{{\text{l}}_{\text{2}}}\]

From this balanced reaction, we can say that 1 mol of \[{\text{Mn}}{{\text{O}}_{\text{2}}}\] produce 1 mol of \[{\text{C}}{{\text{l}}_{\text{2}}}\] gas. So, we can say that 0.05 mol of \[{\text{Mn}}{{\text{O}}_{\text{2}}}\] are needed to produce 0.05 mol of \[{\text{C}}{{\text{l}}_{\text{2}}}\] gas.

 Now, we will calculate the mass \[{\text{Mn}}{{\text{O}}_{\text{2}}}\] needed using its moles and molar mass as follows:
\[{\text{mass = mole }} \times {\text{ molar mass}}\]
The molar mass of \[{\text{Mn}}{{\text{O}}_{\text{2}}}\] = \[8{\text{7}}{\text{.0 g/mol}}\]
\[{\text{mass Mn}}{{\text{O}}_{\text{2}}}{\text{ = }}0.05{\text{ mol Mn}}{{\text{O}}_{\text{2}}} \times 8{\text{7}}{\text{.0 g/mol = 4}}{\text{.35 g}}\]
So, we can say that 4.35 g of \[{\text{Mn}}{{\text{O}}_{\text{2}}}\] needed to produce 1.12L \[{\text{C}}{{\text{l}}_{\text{2}}}\] gas at STP
Hence, x = 4.35g
So,\[{\text{100 x = 100}} \times {\text{ 4}}{\text{.35 = 435}}\].The value of 100x is 435.

Hence, the correct option is (A) 435

Note: Here we have given specific gravity and weight % of HCl solution. Using these values we can calculate the volume of \[{\text{HCl}}\], however, in the question, the only value of 100x is asked to calculate. To solve these types of problems it is very important to balance the reaction as the final answer depends on the stoichiometric ratio of reacting species.