# Question #08af2

Jul 27, 2017

54.375 g

#### Explanation:

$M n {O}_{2} + 4 H C l \to M n C {l}_{2} + 2 {H}_{2} O + C {l}_{2}$

According to the equation,

4 mol of $H C l$ reacts with 87 g of $M n {O}_{2}$
So, 2.5 mol of $H C l$ reacts with $\frac{87 \cdot 2.5}{4}$ g of $M n {O}_{2}$
=54.375 g of $M n {O}_{2}$

Jul 27, 2017

$\text{Mass of} \textcolor{w h i t e}{x} M n {O}_{2} = 36 g$

#### Explanation:

$M n {O}_{2} + 4 H C l \to M n C {l}_{2} + 2 {H}_{2} O + C {l}_{2}$

We are only considering $M n {O}_{2} + 4 H C l$ since the question deals only with the reactants and not the products..

Stoichiometric ratio $\to$ $1 : 4$

Since $1 \text{mole}$ of $M n {O}_{2}$ reacts with $4 \text{moles}$ of $H C l$

Therefore $x$ moles of $M n {O}_{2}$ will react with 2.5 moles of $H C l$

$\Rightarrow$ $\frac{1}{x} = \frac{4}{2.5}$

$\Rightarrow$ $4 \times x = 2.5 \times 1$

$\Rightarrow$ $4 x = 2.5$

$\Rightarrow$ $x = \frac{2.5}{4}$

$\Rightarrow$ $x = 0.625$

Hence no of moles of $M n {O}_{2} = 0.625 m o l s$

Recall $\to$ $\text{No of moles" = "Mass"/"Molar mass}$

$\Rightarrow$ $\text{Mass" = "No of moles" xx "Molar Mass}$

$\text{Molar Mass of} \textcolor{w h i t e}{x} M n {O}_{2} = 25 + \left(16 \times 2\right)$

$\text{Molar Mass of} \textcolor{w h i t e}{x} M n {O}_{2} = 25 + 32$

$\text{Molar Mass of} \textcolor{w h i t e}{x} M n {O}_{2} = 57 g m o {l}^{-} 1$

$\text{Mass of" color(white)(x) MnO_2= "No of moles of" color(white)(x) MnO_2 xx "Molar Mass of} \textcolor{w h i t e}{x} M n {O}_{2}$

$\text{Mass of} \textcolor{w h i t e}{x} M n {O}_{2} = 0.625 m o l s \times 57 g m o {l}^{-} 1$

$\text{Mass of} \textcolor{w h i t e}{x} M n {O}_{2} = 0.625 \cancel{m o l s} \times 57 g \cancel{m o {l}^{-} 1}$

$\text{Mass of} \textcolor{w h i t e}{x} M n {O}_{2} = 35.625 g$

$\text{Mass of} \textcolor{w h i t e}{x} M n {O}_{2} \approx 36 g$