What minimum volume of .2M Na2CO3 is needed to precipitate all the Sr 2+ from 25mL of .1M Sr(NO3)2
25ml/2= 12.5 ml
To find the minimum volume of 0.2 M Na2CO3 needed to precipitate all the Sr2+ from 25 mL of 0.1 M Sr(NO3)2, we need to determine the stoichiometry of the reaction between Na2CO3 and Sr2+.
The balanced chemical equation for the reaction between Na2CO3 and Sr2+ is:
Na2CO3 + Sr(NO3)2 -> SrCO3 + 2NaNO3
From the equation, we can see that 1 mole of Na2CO3 reacts with 1 mole of Sr(NO3)2 to form 1 mole of SrCO3.
To calculate the minimum volume of 0.2 M Na2CO3, we need to determine the number of moles of Sr(NO3)2 in 25 mL of 0.1 M Sr(NO3)2 first.
The number of moles can be calculated using the formula: moles = concentration (M) x volume (L)
Moles of Sr(NO3)2 = 0.1 M x 0.025 L = 0.0025 moles
Since the stoichiometry of the reaction is 1:1, we need the same number of moles of Na2CO3 to react with Sr(NO3)2.
Therefore, we need 0.0025 moles of Na2CO3.
To find the volume of 0.2 M Na2CO3 needed, we can use the formula: volume (L) = moles / concentration (M)
Volume of 0.2 M Na2CO3 = 0.0025 moles / 0.2 M = 0.0125 L = 12.5 mL
Therefore, the minimum volume of 0.2 M Na2CO3 needed to precipitate all the Sr2+ from 25 mL of 0.1 M Sr(NO3)2 is 12.5 mL.