If 45.5 mL of 0.150 M sodium sulfate solution reacts completely with aqueous barium nitrate, what is the mass of #BaSO_4# (233.40 g/mol) precipitate? #Ba(NO_3)_2(aq) + Na_2SO4 (aq) -> BaSO_4(s) + 2NaNO_3(aq)#?

1 Answer
Oct 1, 2016

Answer:

1.59g (3 s.f.)

Explanation:

In this reaction, every 1 mole of #Na_2##SO_4# produces 1 mole of #BaSO_4#, as shown by the mole ratio in the equation. As the question states that all the #Na_2##SO_4# solution reacts, we can assume that #Ba(NO_3)_2# is in excess as it allows for all the #Na_2##SO_4# to react. This means that the reaction is dependent on the moles of #Na_2##SO_4#.

  1. Calculate the moles of #Na_2##SO_4#
    n(#Na_2##SO_4#) = 0.0455L x #(0.150mol)/L# = 0.006825 mol
    Notice that the L units cancel out, leaving only moles; this is basic conversion
  2. Use the mole ratio of #Na_2##SO_4# to #BaSO_4# to find the moles of #BaSO_4#. We established at the beginning that for every 1 mole of #Na_2##SO_4# that reacts, 1 mole of #BaSO_4# is produced. This means that the number of moles of #Na_2##SO_4# that we have will be identical to the number of moles of #BaSO_4# produced.
    n(#BaSO_4#) = n(#Na_2##SO_4#) = 0.006825 mol
  3. Use the molar mass of #BaSO_4# and the moles of #BaSO_4# to calculate the mass of #BaSO_4# produced
    molar mass: 137.3 + 32.07 + 4(16) = 233.37g/mol
    mass produced: 0.006825 mol x #(233.37g)/(mol)# = 1.59g (3 s.f.)