What volume of 0.174 M Na3PO4 solution is necessary to completely react with 89.1 mL of 0.106 M CuCl2?
Here is a simple stoichiometry problem I've posted. Just follow the steps. For moles, remember moles = M x L.
http://www.jiskha.com/science/chemistry/stoichiometry.html
Thank you, I really appreciate
To determine the volume of Na3PO4 solution needed to react with CuCl2, we first need to find the moles of CuCl2 and Na3PO4.
Step 1: Calculate the moles of CuCl2
Moles = Concentration * Volume (in liters)
Given:
Concentration of CuCl2 = 0.106 M
Volume of CuCl2 = 89.1 mL = 0.0891 L
Moles of CuCl2 = 0.106 M * 0.0891 L = 0.00946 moles
Step 2: Determine the stoichiometry between CuCl2 and Na3PO4
The balanced equation for the reaction between CuCl2 and Na3PO4 is:
3 CuCl2 + 2 Na3PO4 -> Cu3(PO4)2 + 6 NaCl
From the equation, we can see that 3 moles of CuCl2 react with 2 moles of Na3PO4.
Step 3: Calculate the volume of Na3PO4 solution needed.
To find the volume, we need to use the stoichiometry ratio:
3 moles CuCl2 = 2 moles Na3PO4
0.00946 moles CuCl2 = (0.00946 moles CuCl2 * 2 moles Na3PO4) / 3 moles CuCl2 = 0.00631 moles Na3PO4
Now we can calculate the volume of Na3PO4 using its concentration:
Concentration of Na3PO4 = 0.174 M
Moles of Na3PO4 = 0.00631 moles
Volume of Na3PO4 = Moles / Concentration = 0.00631 moles / 0.174 M ≈ 0.0362 L
Finally, convert the volume to milliliters:
Volume of Na3PO4 = 0.0362 L * 1000 mL/L ≈ 36.2 mL
Therefore, approximately 36.2 mL of 0.174 M Na3PO4 solution is necessary to completely react with 89.1 mL of 0.106 M CuCl2.
To determine the volume of the Na3PO4 solution needed to react with the CuCl2 solution, we can use the concept of stoichiometry and the balanced chemical equation for the reaction between Na3PO4 and CuCl2.
The balanced chemical equation for the reaction is:
2 Na3PO4 + 3 CuCl2 -> Cu3(PO4)2 + 6 NaCl
From the equation, we can see that 2 moles of Na3PO4 react with 3 moles of CuCl2.
First, we need to calculate the number of moles of CuCl2 in the given volume (89.1 mL) of 0.106 M CuCl2.
Moles of CuCl2 = (Volume in liters) x (Concentration in M)
= 0.0891 L x 0.106 mol/L
= 0.0094466 mol
According to the stoichiometry, 2 moles of Na3PO4 react with 3 moles of CuCl2.
Therefore, we can set up a proportion to calculate the required volume of Na3PO4 solution:
(0.0094466 mol CuCl2 / 3 mol CuCl2) = (Volume of Na3PO4 solution / 0.174 M Na3PO4)
Rearranging the equation, we can solve for the volume of Na3PO4 solution:
Volume of Na3PO4 solution = (0.0094466 mol CuCl2 / 3 mol CuCl2) x (1 L / 0.174 mol)
Volume of Na3PO4 solution = 0.051 L or 51 mL (rounded to 3 significant figures)
Therefore, approximately 51 mL of the 0.174 M Na3PO4 solution is needed to completely react with 89.1 mL of the 0.106 M CuCl2 solution.