What volume of .300 M CaCl2 is needed to react completely with 40.00 Ml of .200 M H2PO4. according to the following equation
3 CaCl2 (aq) + 2 H3PO4 (aq)-> Ca3(po44)2 + 6 HCL (aq)
you need 3/2 moles of CaCl2 for each mole of H3PO4.
moles of phosphoric acid: .04*.2
moles of calcium chloride: 3/2 * .008
moles of calcium chloride= volume*.3
volume= (3/2 .008)/.3 liters
If the 3 moles of Calcium Nitride was then dissolved in 2 L of liquid ammonium what would the molarity of the Calcium Nitride /ammonium solution be?
To determine the volume of 0.300 M CaCl2 needed to react completely with 40.00 mL of 0.200 M H2PO4, we can use the stoichiometry of the balanced equation.
From the balanced equation:
3 CaCl2 (aq) + 2 H3PO4 (aq) -> Ca3(PO4)2 (s) + 6 HCl (aq)
The stoichiometric ratio between CaCl2 and H3PO4 is 3:2. This means that for every 3 moles of CaCl2, we need 2 moles of H3PO4.
1. Convert the given volume of H2PO4 solution in milliliters (mL) to moles:
40.00 mL * (0.200 mol/L) * (1 L/1000 mL) = 0.008 mol H3PO4
2. Use the stoichiometric ratio to determine the moles of CaCl2 required:
2 moles H3PO4 : 3 moles CaCl2
0.008 mol H3PO4 : x mol CaCl2
x = (0.008 mol H3PO4)(3 mol CaCl2 / 2 mol H3PO4) = 0.012 mol CaCl2
3. Convert the moles of CaCl2 to volume in liters (L):
0.012 mol CaCl2 * (1 L/0.300 mol) = 0.040 L
4. Convert the volume from liters to milliliters (mL):
0.040 L * (1000 mL/1 L) = 40.0 mL
Therefore, 40.0 mL of 0.300 M CaCl2 solution is needed to react completely with 40.00 mL of 0.200 M H2PO4 solution.
To determine the volume of 0.300 M CaCl2 required to react completely with 40.00 mL of 0.200 M H2PO4, you need to use stoichiometry and the concept of molarity.
First, let's identify the stoichiometric ratio between CaCl2 and H2PO4 in the balanced equation:
3 CaCl2(aq) + 2 H3PO4(aq) -> Ca3(PO4)2(s) + 6 HCl(aq)
From the equation, we can see that 3 moles of CaCl2 react with 2 moles of H3PO4. This means that the mole ratio of CaCl2 to H3PO4 is 3:2.
Now, let's calculate the number of moles of H2PO4 we have:
Moles of H2PO4 = Molarity × Volume (in liters)
Moles of H2PO4 = 0.200 mol/L × 0.04000 L = 0.00800 moles
Using the mole ratio, we can calculate the moles of CaCl2 required to react with 0.00800 moles of H2PO4:
Moles of CaCl2 = (Moles of H2PO4) × (3 moles of CaCl2 / 2 moles of H3PO4)
Moles of CaCl2 = 0.00800 moles × (3/2) = 0.01200 moles
Next, let's calculate the volume of 0.300 M CaCl2 needed:
Volume of CaCl2 = (Moles of CaCl2) / (Molarity of CaCl2)
Volume of CaCl2 = 0.01200 moles / 0.300 mol/L ≈ 0.04 L = 40.00 mL
Therefore, you would need approximately 40.00 mL of 0.300 M CaCl2 solution to react completely with 40.00 mL of 0.200 M H2PO4.