Question

What would be the minimum weight of ${\text{Mn}}{{\text{O}}_{\text{2}}}$ and minimum volume of HCl of specific gravity 1.2g/ml and 3.65% by weight, needed to produce 1.12 liters of ${\text{C}}{{\text{l}}_2}$ at STP by the reaction give as follows:
${\text{Mn}}{{\text{O}}_{\text{2}}}{\text{ + HCl}} \to {\text{MnC}}{{\text{l}}_{\text{2}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O + C}}{{\text{l}}_{\text{2}}}$
A. 4.35g, 200ml
B. 48.7g, 166.67ml
C. 4.35g, 166.7ml
D. 4.35g, 333.3ml

Answer 1

Hint:Specific gravity or relative density of a substance is the ratio of the density of a substance to a standard substance (mostly water). And the standard temperature and pressure mean a temperature of 273k and 1 atm pressure. Since we are to find the mass and volume of the reactant, the density formula could be of great use.


Complete step by step solution:
Let’s first write the complete balanced equation for the reaction and try to calculate the number of moles of the required reactants and product, the balanced chemical reaction can be written as:
${\text{Mn}}{{\text{O}}_{\text{2}}}{\text{ + 4HCl}} \to {\text{ MnC}}{{\text{l}}_{\text{2}}}{\text{ + 2}}{{\text{H}}_{\text{2}}}{\text{O + C}}{{\text{l}}_{\text{2}}}$
From the above-balanced reaction we conclude the following;
For every 1 mole of ${\text{Mn}}{{\text{O}}_{\text{2}}}$ and 4 moles of HCl, 1 mole of ${\text{C}}{{\text{l}}_{\text{2}}}$ is being produced.
We are given that 1.12L ${\text{C}}{{\text{l}}_{\text{2}}}$ is being produced. This information can give the number of moles of
${\text{C}}{{\text{l}}_{\text{2}}}$ that is being produced.
We know that 1 mole of any substance = 22.4 L
$ \Rightarrow $ 1.12 L of ${\text{C}}{{\text{l}}_{\text{2}}}$ = $\dfrac{1}{{22.4}} \times 1.22$
$ \Rightarrow $ 1.12 L of ${\text{C}}{{\text{l}}_{\text{2}}}$ = 0.05 moles
Now using the conclusion from the balanced chemical equation, we can say the 0.05 mole ${\text{C}}{{\text{l}}_{\text{2}}}$ will be produced by 0.05 moles of ${\text{Mn}}{{\text{O}}_{\text{2}}}$
$ \Rightarrow $ Mass of ${\text{Mn}}{{\text{O}}_{\text{2}}}$ = 0.05 $ \times $86.9 [ since, Given mass = ${\text{No}}{\text{. of moles }} \times {\text{ molar mass}}$, and the molecular mass of ${\text{Mn}}{{\text{O}}_{\text{2}}}$ is 86.9 g/mol ]
$ \Rightarrow $ Mass of ${\text{Mn}}{{\text{O}}_{\text{2}}}$ = 4.35 g
Now, we have to calculate the minimum volume of HCl of 1.2g/ml specific gravity (or density) and 3.65% by weight,
$ \Rightarrow $ 1ml of the solution contains 0.0438g of HCl, (since , 1.2 $ \times $$\dfrac{{3.65}}{{100}}$ = 0.0438g)
$ \Rightarrow $ No. of moles present in 1 ml solution = $\dfrac{{{\text{given mass}}}}{{{\text{molar mass}}}}$ = $\dfrac{{0.0438}}{{36.5}}$ = 0.0012 = 1.2 $ \times $ ${10^{ - 3}}$ mol/ml
Again using the same conclusion by the balanced reaction, we calculate the no. of moles of HCl,
$ \Rightarrow $ No. of moles of HCl = 4 $ \times $0.05 = 0.2mol
$\therefore $ the volume of HCl = $\dfrac{{0.2}}{{1.2 \times {{10}^{ - 3}}}}$
$ \Rightarrow $ The volume of HCl = 166.7 ml
Therefore, the minimum mass ${\text{Mn}}{{\text{O}}_{\text{2}}}$ is 4.35 g, and the minimum volume of HCl is 166.7ml
Hence, the correct answer option (C) i.e., 4.35 g and 166.7 ml.


Note:While calculating the volume be careful of the unit, as most of the time we use the unit in liters, but in this case, we have used ml (mililitre). Also, the unit of volume ${\text{C}}{{\text{l}}_{\text{2}}}$ is given in liters and we did not change the unit as we need the quantity in liters due to the relation, 1 mol of any substance at STP = 22.4 L.