3Ca(NO3)2(aq)+2Na3PO4(aq)->Ca3(PO4)2(s)+6NaNO3(aq)
(a) the grams of Ca3(PO4)2 that can be obtained from 126 mL of 0.750 M Ca(NO3)2
(b) the volume of 0.10 M Na3PO4 needed to react with 13 mL of 0.30 M Ca3(NO3)2
(c) the molarity (M) of the Ca(NO3)2 solution when 46.0 mL react with 57.3 mL of 3.8 M Na3PO4
To find the answers to these questions, we need to use stoichiometry and molarity calculations. Let's go through each question step by step:
(a) To determine the grams of Ca3(PO4)2 that can be obtained from 126 mL of 0.750 M Ca(NO3)2, we first need to calculate the number of moles of Ca(NO3)2 present in the solution.
Step 1: Calculate moles of Ca(NO3)2
Moles = Volume (Liters) x Molarity
Volume in liters = 126 mL / 1000 mL/L = 0.126 L
Moles = 0.126 L x 0.750 mol/L = 0.0945 moles
Step 2: Determine the stoichiometry of the reaction
From the balanced equation, we see that the coefficient of Ca(NO3)2 is 3 and the coefficient of Ca3(PO4)2 is 1. Therefore, the mole ratio between Ca(NO3)2 and Ca3(PO4)2 is 3:1.
Step 3: Convert moles of Ca(NO3)2 to moles of Ca3(PO4)2
Moles of Ca3(PO4)2 = Moles of Ca(NO3)2 x (1 mole Ca3(PO4)2 / 3 moles Ca(NO3)2)
Moles of Ca3(PO4)2 = 0.0945 moles x (1/3) = 0.0315 moles
Step 4: Calculate the molecular weight of Ca3(PO4)2
Ca3(PO4)2 consists of three calcium (Ca) atoms with a molar mass of 40.08 g/mol, two phosphate (PO4) groups with a molar mass of 94.97 g/mol, and a total of ten oxygen (O) atoms with a molar mass of 16.00 g/mol.
Molecular weight of Ca3(PO4)2 = (3 x 40.08 g/mol) + (2 x 94.97 g/mol) + (10 x 16.00 g/mol) = 310.18 g/mol
Step 5: Calculate the grams of Ca3(PO4)2
Grams of Ca3(PO4)2 = Moles x Molecular weight
Grams of Ca3(PO4)2 = 0.0315 moles x 310.18 g/mol = 9.81 grams
Therefore, 9.81 grams of Ca3(PO4)2 can be obtained from 126 mL of 0.750 M Ca(NO3)2.
(b) To find the volume of 0.10 M Na3PO4 needed to react with 13 mL of 0.30 M Ca3(NO3)2, we can use the stoichiometry and molarity to determine the number of moles of each reactant.
Step 1: Calculate moles of Ca3(NO3)2
Moles = Volume (Liters) x Molarity
Volume in liters = 13 mL / 1000 mL/L = 0.013 L
Moles of Ca3(NO3)2 = 0.013 L x 0.30 mol/L = 0.0039 moles
Step 2: Determine the stoichiometry of the reaction
From the balanced equation, we see that the coefficient of Ca3(NO3)2 is 1 and the coefficient of Na3PO4 is 2. Therefore, the mole ratio between Ca3(NO3)2 and Na3PO4 is 1:2.
Step 3: Convert moles of Ca3(NO3)2 to moles of Na3PO4
Moles of Na3PO4 = Moles of Ca3(NO3)2 x (2 moles Na3PO4 / 1 mole Ca3(NO3)2)
Moles of Na3PO4 = 0.0039 moles x 2 = 0.0078 moles
Step 4: Calculate the volume of 0.10 M Na3PO4
Volume (Liters) = Moles / Molarity
Volume (Liters) = 0.0078 moles / 0.10 mol/L = 0.078 L
Therefore, 0.078 liters or 78 mL of 0.10 M Na3PO4 is needed to react with 13 mL of 0.30 M Ca3(NO3)2.
(c) To determine the molarity of the Ca(NO3)2 solution when 46.0 mL react with 57.3 mL of 3.8 M Na3PO4, we can use the stoichiometry and molarity to calculate the number of moles and then the molarity.
Step 1: Calculate moles of Na3PO4
Moles = Volume (Liters) x Molarity
Volume in liters for Na3PO4 = 57.3 mL / 1000 mL/L = 0.0573 L
Moles of Na3PO4 = 0.0573 L x 3.8 mol/L = 0.21774 moles
Step 2: Determine the stoichiometry of the reaction
From the balanced equation, we see that the coefficient of Na3PO4 is 2 and the coefficient of Ca(NO3)2 is 3. Therefore, the mole ratio between Na3PO4 and Ca(NO3)2 is 2:3.
Step 3: Convert moles of Na3PO4 to moles of Ca(NO3)2
Moles of Ca(NO3)2 = Moles of Na3PO4 x (3 moles Ca(NO3)2 / 2 moles Na3PO4)
Moles of Ca(NO3)2 = 0.21774 moles x (3/2) = 0.3266 moles
Step 4: Calculate the molarity of Ca(NO3)2
Molarity = Moles / Volume (Liters)
Volume in liters for Ca(NO3)2 = 46.0 mL / 1000 mL/L = 0.046 L
Molarity = 0.3266 moles / 0.046 L = 7.105 M
Therefore, the molarity of the Ca(NO3)2 solution is approximately 7.105 M when 46.0 mL reacts with 57.3 mL of 3.8 M Na3PO4.