2. For the addition of 125.00 mL of 0.1352 M calcium bromide to 175.00 mL of 0.1

Question

2. For the addition of 125.00 mL of 0.1352 M calcium bromide to 175.00 mL of 0.1015 M sodium oxalate, determine the following: a. Write the balanced molecular equation for the reaction.b. What is the limiting reagent? c. What is the molarity of all ions in the final solution? d. Assuming the reaction proceeds at 100 %, what volume of the limiting reagent is required to produce 45.50 g of the precipitate if the concentrations remain the same? e. What molarity of the limiting reagent would be required if 100.00 mL of that solution were used and the desired amount of precipitate was 75.00 g?

Explanation / Answer

Balanced equation:
CaBr2 + Na2C2O4 ====> CaC2O4 + 2 NaBr
Reaction type: double replacement

Moles of CaBr2 = 125 x 0.1352 /1000 = 0.0169 Moles

Moles of Soidum Oxalate = 175 x 0.1015 /1000 = 0.0177625 Moles

One equivalent of each reagent is needed

Hence Limiting reagent is Calcium Bromide

45.5 gm of calcium oxalte = 45.5 / 128.09 =  0.35519 Moles

Volume of the solution needed = 0.35519 x 1000 / 0.1352 = 2627 ml or 2.627 Liter

75 gm of calcium oxalte = 75 / 128.09 =  0.5854 Moles

Volume of the solution needed = 0.5854 x 1000 / 0.1352 = 4329 ml or 4.329 Liter

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