Tyler S. asked • 04/26/16
Balanced equation: 2Na3PO4(ac) + 3Cu(NO3)2(ac) = Cu3(PO4)2(s) + 6NaNO3(ac) Na3PO4 molarity: (1.0 M) Cu(NO3)2: (5.0 M) I don't know how to calculate how much of 2Na3PO4(ac) and 3Cu(NO3)2(ac) is required (in mL) to make 10 grams of Cu3(PO4)2. 2 Answers By Expert Tutors Na3PO4 molarity: (1.0 M) Cu(NO3)2: (5.0 M) STEP 1. , Find the molar mass of the following (use a Periodic Table): Cu3(PO4)2 =3(63.55) + 2(31 +64) = 380.65 STEP 2. Change the grams to moles. moles = grams/ molar mass 10 g Cu3(PO4)2 = 10/380.65 = 0.0263 mol STEP 3. Use the mole ratios. By the mole ratios (look at the coefficients from the equation), it requires 3 moles Cu(NO3)2 and 2 moles Na3PO4 to form 1 mole of Cu3(PO4)2 This means that to produce 0.0263 mole Cu3(PO4)2, 3 x 0.0263 = 0.0789 mole Cu(NO3)2 and 2 x 0.0263 = 0.0526 mol Na3PO4 are required. STEP 4. Use the Molarity formula to find the volume. or Liters = moles/ Molarity Volume of Cu(NO3)2 = 0.0789 mol/ 5.0 M = 0.01578 L ≈ 16 mL Volume of Na3PO4 = 0.0526 mol/ 1.0 M = 0.0526 L ≈ 53 mL
Jacoby B. answered • 04/26/16 Math, Science and Test Prep Tutor
I see you have the molarity for the two solutions but you would need a bit more information like the starting mass from one of the products or liters of solution. |