Respuesta :
Answer :
The mass of excess mass of [tex]Na_2CO_3[/tex], [tex]AgNO_3,Ag_2CO_3\text{ and }NaNO_3[/tex] are, 1.908 g, 0 g, 12.144 g and 3.74 g respectively.
Explanation : Given,
Mass of [tex]Na_2CO_3[/tex] = 4.25 g
Mass of [tex]AgNO_3[/tex] = 7.50 g
Molar mass of [tex]Na_2CO_3[/tex] = 106 g/mole
Molar mass of [tex]AgNO_3[/tex] = 170 g/mole
Molar mass of [tex]Ag_2CO_3[/tex] = 276 g/mole
Molar mass of [tex]NaNO_3[/tex] = 85 g/mole
First we have to calculate the moles of [tex]Na_2CO_3[/tex] and [tex]AgNO_3[/tex].
[tex]\text{Moles of }Na_2CO_3=\frac{\text{Mass of }Na_2CO_3}{\text{Molar mass of }Na_2CO_3}=\frac{4.25g}{106g/mole}=0.040moles[/tex]
[tex]\text{Moles of }AgNO_3=\frac{\text{Mass of }AgNO_3}{\text{Molar mass of }AgNO_3}=\frac{7.50g}{170g/mole}=0.044moles[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
[tex]Na_2CO_3+2AgNO_3\rightarrow Ag_2CO_3+2NaNO_3[/tex]
From the balanced reaction we conclude that
As, 2 moles of [tex]AgNO_3[/tex] react with 1 mole of [tex]Na_2CO_3[/tex]
So, 0.044 moles of [tex]AgNO_3[/tex] react with [tex]\frac{0.044}{2}=0.022[/tex] moles of [tex]Na_2CO_3[/tex]
From this we conclude that, [tex]Na_2CO_3[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]AgNO_3[/tex] is a limiting reagent and it limits the formation of product.
The excess mole of [tex]Na_2CO_3[/tex] = 0.040 - 0.022 = 0.018 mole
Now we have to calculate the mass of excess mole of [tex]Na_2CO_3[/tex].
[tex]\text{Mass of }Na_2CO_3=\text{Moles of }Na_2CO_3\times \text{Molar mass of }Na_2CO_3=(0.018mole)\times (106g/mole)=1.908g[/tex]
Now we have to calculate the moles of [tex]Ag_2CO_3[/tex].
As, 1 moles of [tex]AgNO_3[/tex] react to give 1 moles of [tex]Ag_2CO_3[/tex]
So, 0.044 moles of [tex]AgNO_3[/tex] react to give 0.044 moles of [tex]Ag_2CO_3[/tex]
Now we have to calculate the mass of [tex]AgCO_3[/tex].
[tex]\text{Mass of }Ag_2CO_3=\text{Moles of }Ag_2CO_3\times \text{Molar mass of }Ag_2CO_3=(0.044mole)\times (276g/mole)=12.144g[/tex]
Now we have to calculate the moles of [tex]NaNO_3[/tex].
As, 2 moles of [tex]AgNO_3[/tex] react to give 2 moles of [tex]NaNO_3[/tex]
So, 0.044 moles of [tex]AgNO_3[/tex] react to give 0.044 moles of [tex]NaNO_3[/tex]
Now we have to calculate the mass of [tex]NaNO_3[/tex].
[tex]\text{Mass of }NaNO_3=\text{Moles of }NaNO_3\times \text{Molar mass of }NaNO_3=(0.044mole)\times (85g/mole)=3.74g[/tex]
