Question

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)`?

Answer 1

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.)