what mass of Ca is needed to completely react with 5.0M of .250ml Copper I Phosphate?
To determine the mass of calcium needed to completely react with copper(I) phosphate, we need to set up and balance the chemical equation and use stoichiometry to find the molar ratio between calcium and copper(I) phosphate.
The balanced chemical equation for the reaction between calcium and copper(I) phosphate (Cu3PO4) can be written as:
3Ca + 2Cu3PO4 → 6CaPO4 + Cu2P
From the balanced equation, we can see that the molar ratio between calcium and copper(I) phosphate is 3:2.
Step 1: Convert the volume of copper(I) phosphate solution (0.250 mL) to moles.
Since the concentration of the copper(I) phosphate solution is given as 5.0 M (moles per liter), we can calculate the number of moles of copper(I) phosphate in 0.250 mL as follows:
0.250 mL = 0.250/1000 = 0.00025 L (conversion from mL to L)
Number of moles of copper(I) phosphate = concentration × volume
= 5.0 mol/L × 0.00025 L
= 0.00125 mol
Step 2: Use the molar ratio from the balanced equation to find the moles of calcium needed.
According to the balanced equation, the molar ratio between calcium and copper(I) phosphate is 3:2. Therefore, for every 2 moles of copper(I) phosphate, we need 3 moles of calcium.
Number of moles of calcium = (3/2) × number of moles of copper(I) phosphate
= (3/2) × 0.00125 mol
= 0.001875 mol
Step 3: Calculate the mass of calcium needed using its molar mass.
The molar mass of calcium (Ca) is approximately 40.08 g/mol.
Mass of calcium = number of moles × molar mass
= 0.001875 mol × 40.08 g/mol
= 0.0752 g
Therefore, approximately 0.0752 grams of calcium are needed to completely react with 5.0 M, 0.250 mL of copper(I) phosphate solution.