Respuesta :
Answer : The HCl reactant is left over in the reaction and the left over amount of HCl is, 3.1 grams
Explanation : Given,
Mass of [tex]MnO_2[/tex] = 16.0 g
Mass of [tex]HCl[/tex] = 30.0 g
Molar mass of [tex]MnO_2[/tex] = 87 g/mole
Molar mass of [tex]HCl[/tex] = 36.5 g/mole
First we have to calculate the moles of [tex]MnO_2[/tex] and [tex]HCl[/tex].
[tex]\text{Moles of }MnO_2=\frac{\text{Mass of }MnO_2}{\text{Molar mass of }MnO_2}=\frac{16g}{87g/mole}=0.184moles[/tex]
[tex]\text{Moles of }HCl=\frac{\text{Mass of }HCl}{\text{Molar mass of }HCl}=\frac{30g}{36.5g/mole}=0.822moles[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
[tex]MnO_2+4HCl\rightarrow MnCl_2+Cl_2[/tex]
From the balanced reaction we conclude that
As, 1 moles of [tex]MnO_2[/tex] react with 4 mole of [tex]HCl[/tex]
So, 0.184 moles of [tex]MnO_2[/tex] react with [tex]4\times 0.184=0.736[/tex] moles of [tex]HCl[/tex]
From this we conclude that, [tex]HCl[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]MnO_2[/tex] is a limiting reagent and it limits the formation of product.
The excess moles of HCl = 0.822 - 0.736 = 0.086 mole
Now we have to calculate the mass of [tex]HCl[/tex].
[tex]\text{Mass of }HCl=\text{Moles of }HCl\times \text{Molar mass of }HCl[/tex]
[tex]\text{Mass of }HCl=(0.086mole)\times (36.5g/mole)=3.139g\approx 3.1g[/tex]
Therefore, the HCl reactant is left over in the reaction and the left over amount of HCl is, 3.1 grams
