Respuesta :
If the volume of the solution is 2.0 liters and the concentration of NO3 - is 0.20 m, the number of mol of NO3- in the solution would be:
mol= volume * concentration= 2 liters * 0.2 mol/liter= 0.4 mol
The equation of ion formed by ca(no 3) 2 would be:
Ca(NO3)2 ==> 1 Ca + 2 NO3
For every 1 mol of Ca(NO3)2 will produce 2 mol of NO3-. So, the amount of Ca(NO3)2 needed to make 0.4 mol of NO3- would be: 0.4 mol / (2/1)= 0.2 mol
mol= volume * concentration= 2 liters * 0.2 mol/liter= 0.4 mol
The equation of ion formed by ca(no 3) 2 would be:
Ca(NO3)2 ==> 1 Ca + 2 NO3
For every 1 mol of Ca(NO3)2 will produce 2 mol of NO3-. So, the amount of Ca(NO3)2 needed to make 0.4 mol of NO3- would be: 0.4 mol / (2/1)= 0.2 mol
Answer : The number of moles of [tex]Ca(NO_3)_2[/tex] required are, 0.2 moles
Explanation :
First we have to calculate the moles of [tex]NO_3^-[/tex].
[tex]\text{Moles of }NO_3^-=\text{Molarity of }NO_3^-\times \text{Volume of solution}=0.20M\times 2.0L=0.4mole[/tex]
Now we have to calculate the moles of [tex]Ca(NO_3)_2[/tex].
The dissociation chemical reaction will be,
[tex]Ca(NO_3)_2\rightarrow Ca^{2+}+2NO_3^-[/tex]
From this reaction we conclude that,
As, 2 moles of [tex]NO_3^-[/tex] obtained from 1 mole of [tex]Ca(NO_3)_2[/tex]
So, 0.4 moles of [tex]NO_3^-[/tex] obtained from [tex]\frac{0.4}{2}=0.2[/tex] mole of [tex]Ca(NO_3)_2[/tex]
Therefore, the number of moles of [tex]Ca(NO_3)_2[/tex] required are, 0.2 moles
