Use your balanced equation to calculate the volume of 0.12 M Na3PO4 to stoichiometrically react with the 3.0 mL of Co (NO3)2. Show your work.
You don't have enough information.
mols Na3PO4 = M x L but I don't see a L.
Same for the Co salt, no M
To calculate the volume of 0.12 M Na3PO4 needed to react with 3.0 mL of Co(NO3)2, we need to use the balanced equation and the stoichiometry of the reaction.
The balanced equation for the reaction is:
2 Co(NO3)2 + 3 Na3PO4 -> Co3(PO4)2 + 6 NaNO3
From the balanced equation, we can see that the stoichiometric ratio between Co(NO3)2 and Na3PO4 is 2:3.
First, let's convert the volume of Co(NO3)2 to moles using its molarity:
Molarity = moles / volume
Rearranging the equation:
moles = Molarity × volume
Given that the volume of Co(NO3)2 is 3.0 mL (which is the same as 0.003 L) and assuming the molarity of Co(NO3)2 is also 0.12 M, we can calculate the moles of Co(NO3)2:
moles of Co(NO3)2 = 0.12 M × 0.003 L = 0.00036 moles
Since the stoichiometric ratio between Co(NO3)2 and Na3PO4 is 2:3, the moles of Na3PO4 needed will be:
moles of Na3PO4 = moles of Co(NO3)2 × (3/2) = 0.00036 moles × (3/2) = 0.00054 moles
Finally, let's convert the moles of Na3PO4 to volume using its molarity:
Volume = moles / Molarity
Given that the molarity of Na3PO4 is 0.12 M, the volume of Na3PO4 needed will be:
Volume of Na3PO4 = 0.00054 moles / 0.12 M = 0.0045 L
Converting the volume to milliliters:
Volume of Na3PO4 = 0.0045 L × 1000 mL/L = 4.5 mL
Therefore, the volume of 0.12 M Na3PO4 needed to stoichiometrically react with 3.0 mL of Co(NO3)2 is 4.5 mL.
To calculate the volume of 0.12 M Na3PO4 needed to stoichiometrically react with 3.0 mL of Co(NO3)2, we first need to write a balanced equation for the reaction between Na3PO4 and Co(NO3)2:
3 Na3PO4 + 2 Co(NO3)2 -> Co3(PO4)2 + 6 NaNO3
From the balanced equation, we can see that the molar ratio between Na3PO4 and Co(NO3)2 is 3:2. This means that for every 3 moles of Na3PO4, we need 2 moles of Co(NO3)2.
Given that the concentration of Na3PO4 is 0.12 M, we can convert this to moles per liter (Molarity = moles/volume in liters):
0.12 M Na3PO4 = 0.12 moles/L
Now, we need to determine the number of moles of Co(NO3)2 in the given volume of 3.0 mL. To do this, we'll perform a conversion using the molar concentration of Co(NO3)2:
Moles of Co(NO3)2 = Concentration x Volume
Moles of Co(NO3)2 = 0.12 M x (3.0 mL / 1000 mL/L) = 0.00036 moles
Since the molar ratio between Na3PO4 and Co(NO3)2 is 3:2, we can determine the number of moles of Na3PO4 needed to react with 0.00036 moles of Co(NO3)2:
Moles of Na3PO4 = (2/3) x Moles of Co(NO3)2
Moles of Na3PO4 = (2/3) x 0.00036 moles = 0.00024 moles
Finally, we can determine the volume of 0.12 M Na3PO4 needed to achieve 0.00024 moles:
Volume = Moles / Concentration
Volume = 0.00024 moles / 0.12 M = 0.002 L = 2.0 mL
Therefore, 2.0 mL of 0.12 M Na3PO4 is needed to stoichiometrically react with 3.0 mL of Co(NO3)2.