Question

Use the following reaction from the lab to calculate the following: 2na3po4 • 12h2o(s) 3bacl2 • 2h2o(s) → ba3(po4)2(s) 6naci(aq) 30h20(/)

a. if 2 grams of ba3(po4)2(s) (precipitate) are made, how many grams of bacl2 • 2h2o did you start with?
b. if 1.5 grams of ba3(po4)2(s) (precipitate) are made, how many grams of 2na3po4 • 12h20 did you start with?

Answer 1

To solve this problem, we need to use the balanced chemical equation for the reaction and apply stoichiometry to calculate the grams of reactants. For part a, if 2 grams of Ba3(PO4)2 is made, we calculate the grams of BaCl2.2H2O. For part b, if 1.5 grams of Ba3(PO4)2 is made, we calculate the grams of 2Na3PO4.12H2O.

Step-by-step explanation:

To solve this problem, we need to use the balanced chemical equation for the reaction and apply stoichiometry. From the balanced equation:

• 2 moles of Ba3(PO4)2 correspond to 3 moles of BaCl2.2H2O
• We can convert grams of Ba3(PO4)2 to moles using its molar mass and then use stoichiometry to find the moles of BaCl2.2H2O.
• Finally, we convert moles of BaCl2.2H2O to grams using its molar mass.

For part a, if 2 grams of Ba3(PO4)2 is made, we calculate:

1. Convert grams of Ba3(PO4)2 to moles
2. Use stoichiometry to find moles of BaCl2.2H2O
3. Convert moles of BaCl2.2H2O to grams

For part b, the process is similar but with different values. If 1.5 grams of Ba3(PO4)2 is made, we need to calculate the grams of 2Na3PO4.12H2O.