Respuesta :
Answer : The excess reactant is, [tex]H_2O[/tex]
The leftover amount of excess reagent is, 7.2 grams.
Solution : Given,
Mass of [tex]CaCN_2[/tex] = 105.0 g
Mass of [tex]H_2O[/tex] = 78.0 g
Molar mass of [tex]CaCN_2[/tex] = 80.11 g/mole
Molar mass of [tex]H_2O[/tex] = 18 g/mole
Molar mass of [tex]CaCO_3[/tex] = 100.09 g/mole
First we have to calculate the moles of [tex]CaCN_2[/tex] and [tex]H_2O[/tex].
[tex]\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles[/tex]
[tex]\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
[tex]CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3[/tex]
From the balanced reaction we conclude that
As, 1 mole of [tex]CaCN_2[/tex] react with 3 mole of [tex]H_2O[/tex]
So, 1.31 moles of [tex]CaCN_2[/tex] react with [tex]1.31\times 3=3.93[/tex] moles of [tex]H_2O[/tex]
From this we conclude that, [tex]H_2O[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]CaCN_2[/tex] is a limiting reagent and it limits the formation of product.
Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles
Now we have to calculate the mass of excess reactant.
[tex]\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)[/tex]
[tex]\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g[/tex]
Thus, the leftover amount of excess reagent is, 7.2 grams.
