Question

H3AsO4 + 3KOH → 3H2O + Na3AsO4 11.0 milliliters of 0.316 M H3AsO4 is mixed with 31.6 milliliters of 0.300 M KOH.

What is the molarity of Na3AsO4 in the resulting solution?

BaBr₂(aq) + (NH₄)₂CO₃(aq) → BaCO₃(s) + 2NH₄Br(aq)

If 75.0 milliliters of 0.804 M BaBr₂ is allowed to react with 100.0 milliliters of 0.403 M (NH₄)₂CO₃

how many grams of BaCO₃ will be produced?

Answer 1

has ou + 3 KOH Koz Asou + 340 minores of H₂ AsOy = MX V = 0.316* 11.0 = 3.476 mmol mmoles or KOH= 0.30X316 = 9.48 mmor Kot in limiting mmol of kozasoy tormes= 1x 9.48 3.16 mmus [Ks As Qu] = 3-16 - 10.0742 JM 49.6 - BaBb + (NH4) Coz Baloz + 9 Nhy By moles of Baß = 0.804xo.ots = 0.0603 mol moles of (NH4) Coz= 0.4031 ol = 0.0403 mar moles of Balog Produced = 0.0403 mot man = 0.0403x 197= 11.9q Balon