If 160 mL of 2.75 M aqueous Na 3 PO 4 and830 mL of 0.217 M aqueous CaCl 2 are re

Question

If 160 mL of 2.75 M aqueous Na3PO4 and830 mL of 0.217 M aqueous CaCl2 are reactedstoichiometrically according to the balanced equation, how manymilliliters of 2.35 M aqueous NaCl are produced?

3CaCl2(aq) +2Na3PO4(aq) Ca3(PO4)2(s) +6NaCl(aq)
NO IDEA HOW TO SOLVE. THEPROFESSOR NOR TA HAVE TAUGHT THIS TO US YET.
NO IDEA HOW TO SOLVE. THEPROFESSOR NOR TA HAVE TAUGHT THIS TO US YET.

Explanation / Answer

n(Na3PO4 ) = cv = 2.75 x 0.16 = 0.44mol n(CaCl2 ) = cv = 0.217 x 0.83 = 0.1801mol Stoichiometry ratio = n(Na3PO4 ) /n(CaCl2) = 2/3 = 0.667 Actual ratio = 0.44/0.1801 = 2.44 AR > SR So, CaCl2 is the limiting reagent n(NaCl) = 6/3 x 0.1801 = 0.3602mol c(NaCl) = n/v 2.35 = 0.3602/v V(NaCl) = 0.1533L = 153.3mL Hope this helps!

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