Ask Question
10 January, 07:17

A 50.0 mL sample of 1.54*10-2 M NaSO4 is added to 50.0 mL of 1.28*10-2 Ca (NO3) 2. What percentage of the Ca2 + remains unprecipitated?,

+1
Answers (1)
  1. 10 January, 07:42
    0
    The reaction would be as shown below;

    Na2SO4 + Ca (NO3) 2 = CaSO4 + 2 NaNO3

    The moles of NaSO4 will be;

    = 0.05 * 0.0154 = 0.00077 moles

    While the number of moles of Ca (NO3) 2

    = 0.05 * 0.0128 = 0.00064 moles

    The mole ratio of sodium sulfate and calcium nitrate is 1:1

    Ca (NO3) 2 is the limiting reactant, so ignoring the Ksp of CaSO4, zero percent of the Ca^2 + ions remain unprecipitated.
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “A 50.0 mL sample of 1.54*10-2 M NaSO4 is added to 50.0 mL of 1.28*10-2 Ca (NO3) 2. What percentage of the Ca2 + remains unprecipitated?, ...” in 📗 Chemistry if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers