Question

18) (5 pts) What mass (g) of CaF2 is formed when 478 mL of 0.334 M NaF is treated with an excess of aqueous calcium nitrate DJ 0.623 [AJ 0.943 [BJ 0.472 C] 1.25

Answer 1

Solution Balanud equation !- 2 Naf (aq) + Ca(NO3), laq) laf2 (8) + 2NaNO3(aq) B (excess) If callO3)₂ is in excess to limiting reagent is Naf 0.334 m Naf = moles of Naf 478 me x 1032 - Iml males of NQF = 0.334 moles x 4768X10 | 7TH = 0.0159652 males geren from balanced equation we can see a Naf will form 1 mole of caf2(s) = 0.015965 2 moles of Naf will form = 0.01 596 52 cafe moles = 7,9826 X10-3 moles mole = mass of Cafe molau me melau mans Cafe ->) mass of caf, l . = 78.07g/mol = 7.9826 x10 molx 78'077 = [0,623 g)