# 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)?

Oct 1, 2016

1.59g (3 s.f.)

#### Explanation:

In this reaction, every 1 mole of $N {a}_{2}$$S {O}_{4}$ produces 1 mole of $B a S {O}_{4}$, as shown by the mole ratio in the equation. As the question states that all the $N {a}_{2}$$S {O}_{4}$ solution reacts, we can assume that $B a {\left(N {O}_{3}\right)}_{2}$ is in excess as it allows for all the $N {a}_{2}$$S {O}_{4}$ to react. This means that the reaction is dependent on the moles of $N {a}_{2}$$S {O}_{4}$.

1. Calculate the moles of $N {a}_{2}$$S {O}_{4}$
n($N {a}_{2}$$S {O}_{4}$) = 0.0455L x $\frac{0.150 m o l}{L}$ = 0.006825 mol
Notice that the L units cancel out, leaving only moles; this is basic conversion
2. Use the mole ratio of $N {a}_{2}$$S {O}_{4}$ to $B a S {O}_{4}$ to find the moles of $B a S {O}_{4}$. We established at the beginning that for every 1 mole of $N {a}_{2}$$S {O}_{4}$ that reacts, 1 mole of $B a S {O}_{4}$ is produced. This means that the number of moles of $N {a}_{2}$$S {O}_{4}$ that we have will be identical to the number of moles of $B a S {O}_{4}$ produced.
n($B a S {O}_{4}$) = n($N {a}_{2}$$S {O}_{4}$) = 0.006825 mol
3. Use the molar mass of $B a S {O}_{4}$ and the moles of $B a S {O}_{4}$ to calculate the mass of $B a S {O}_{4}$ produced
molar mass: 137.3 + 32.07 + 4(16) = 233.37g/mol
mass produced: 0.006825 mol x $\frac{233.37 g}{m o l}$ = 1.59g (3 s.f.)