What volume of 0.649 M K3PO4 is required to react with 26 mL of 0.453 M MgCl2 according to the equation
2 K3PO4 + 3 MgCl2 → Mg3(PO4)2 + 6 KCl Answer in units of mL.
I like to work in millimoles to keep the zeros to a minimum.
millimoles = mL x M = ?
2 K3PO4 + 3 MgCl2 → Mg3(PO4)2 + 6 KCl
millimoles MgCl2 = 26 x 0.453 = 11.78
Convert to millimoles K3PO4 using the coefficients in the balanced equation.
11.78 mmols MgCl2 x (2 mols K3PO4/3 mols MgCl2) = 11.78 x 2/3 = 7.85
Then M = millimoles/mL = ? You have M and mmoles, substitute and solve for mL K3PO4.
To determine the volume of 0.649 M K3PO4 required to react with 26 mL of 0.453 M MgCl2, we need to find the stoichiometric ratio between the two reactants.
From the balanced equation:
2 K3PO4 + 3 MgCl2 → Mg3(PO4)2 + 6 KCl
We can see that the ratio of K3PO4 to MgCl2 is 2:3. This means that for every 2 moles of K3PO4, we need 3 moles of MgCl2.
Let's calculate the number of moles of MgCl2 in 26 mL of the solution:
0.453 M = 0.453 moles/L
0.453 moles/L x 0.026 L = 0.011778 moles of MgCl2
According to the stoichiometric ratio, we need 2/3 times the moles of K3PO4. So, the moles of K3PO4 required would be:
(2/3) x 0.011778 = 0.007852 moles of K3PO4
Now, let's calculate the volume of 0.649 M K3PO4 required to have 0.007852 moles:
0.649 M = 0.649 moles/L
Volume (L) = moles/concentration
Volume (L) = 0.007852 moles / 0.649 moles/L
Volume (L) = 0.012099 L
Since 1 L = 1000 mL, we can find the volume in mL:
Volume (mL) = 0.012099 L x 1000 mL/L
Volume (mL) ≈ 12.1 mL
Therefore, the volume of 0.649 M K3PO4 required to react with 26 mL of 0.453 M MgCl2 is approximately 12.1 mL.
To determine the volume of 0.649 M K3PO4 required to react with 26 mL of 0.453 M MgCl2, we first need to calculate the number of moles of MgCl2 present in the 26 mL solution.
Number of moles of MgCl2 = concentration of MgCl2 x volume of MgCl2 solution
= 0.453 M x 0.026 L (since 1 mL = 0.001 L)
= 0.011778 moles
From the balanced chemical equation, we can see that the stoichiometric ratio between K3PO4 and MgCl2 is 2:3.
This means that for every 2 moles of K3PO4, we need 3 moles of MgCl2. Therefore, the number of moles of K3PO4 required would be:
Number of moles of K3PO4 = (3/2) x number of moles of MgCl2
= (3/2) x 0.011778 moles
= 0.017667 moles
Next, we can calculate the volume of 0.649 M K3PO4 needed, using the equation:
Volume of K3PO4 (in liters) = number of moles of K3PO4 / concentration of K3PO4
= 0.017667 moles / 0.649 M
≈ 0.0272 L
Finally, we convert the volume from liters to milliliters:
Volume of K3PO4 (in mL) = 0.0272 L x 1000 mL/L
≈ 27.2 mL
Therefore, approximately 27.2 mL of 0.649 M K3PO4 is required to react with 26 mL of 0.453 M MgCl2.