Question

The reaction of $Mn{O_2}$ and HCl produces 2.24 L of $C{l_2}$ gas. What mass of $Mn{O_2}$ was used up by the reaction? (At. Mass of Mn = $55\dfrac{g}{{mol}}$)
$Mn{O_2} + 4HCl \to 2{H_2}O + MnC{l_2} + C{l_2}$
A. 7.19g
B. 8.7g
C. 2.24g
D. 18.69g

Answer 1

Hint: At first, calculate the molar mass of $Mn{O_2}$ and then use the unitary method to solve the question i.e., molar mass of $Mn{O_2}$ produces a volume of 22.4 L of $C{l_2}$. Then, the amount of mass of $Mn{O_2}$ used up would be calculated on dividing the product of molar mass and volume at S.T.P by the given volume.

Complete Step by step answer:
When one mole of manganese dioxide $\left( {Mn{O_2}} \right)$ reacts with 4 moles of hydrogen chloride $\left( {HCl} \right)$, two moles of water $\left( {{H_2}O} \right)$, one mole of manganese chloride $\left( {MnC{l_2}} \right)$ and one mole of Chlorine gas is formed. The following reaction is shown by the given equation:
 $Mn{O_2} + 4HCl \to 2{H_2}O + MnC{l_2} + C{l_2}$
At standard temperature and pressure i.e., at S.T.P 1 mole of any substance produces a volume of 22.4 L. Thus, 1 mole of $Mn{O_2}$ produces a volume of 22.4 L of $C{l_2}$ as 1 mole of Cl2 is formed from 1 mole of MnO2. Now, we’ll calculate molar mass of $Mn{O_2}$ where molar or atomic mass of Mn is given as 55 $\dfrac{g}{{mol}}$ and atomic mass of oxygen is 16 $\dfrac{g}{{mol}}$.
Molar mass of $Mn{O_2}$ $\left( M \right)$ = 1 $ \times $ atomic mass of Mn $ + $ 2 $ \times $ atomic mass of O.
$M = 1 \times 55 + 2 \times 16$
$\Rightarrow M = 55 + 32$
$\Rightarrow M = 87g$
Thus, the amount of mass of $Mn{O_2}$ required to produce 22.4 L of $C{l_2}$ gas = 87g as 1 mole of MnO2 = 87g.
Amount of mass of $Mn{O_2}$ required to produce 1 L of $C{l_2}$ gas = $\dfrac{{87}}{{22.4}}$.
So, amount of mass of $Mn{O_2}$ required to produce 2.24 L of $C{l_2}$ gas = $\dfrac{{87 \times 2.24}}{{22.4}} i.e.,\dfrac{{87}}{{10}}$.
Hence, amount of mass of $Mn{O_2}$ used up by the reaction to produce 2.24 L of $C{l_2}$ gas is $\dfrac{{87}}{{10}}$ i.e., 8.7g.

Therefore, option B is correct.

Note: Remember that one mole of a substance produces a volume of 22.4 L at S.T.P and one mole is equal to the molar mass of that substance. Also, remember the unitary method and be careful while doing the calculations as both the volumes are the same in digits and they only differ in decimal point.