If 45.5 mL of 0.150 M sodium sulfate solution reacts completely with aqueous barium nitrate, what is the mass of $B a S O_{4}$ $BaSO_4$ (233.40 g/mol) precipitate? $B a {(N O_{3})}_{2} (a q) + N a_{2} S O 4 (a q) \to B a S O_{4} (s) + 2 N a N O_{3} (a q)$ $Ba(NO_3)_2(aq) + Na_2SO4 (aq) -> BaSO_4(s) + 2NaNO_3(aq)$ ?

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If 45.5 mL of 0.150 M sodium sulfate solution reacts completely with aqueous barium nitrate, what is the mass of $B a S O_{4}$ $BaSO_4$ (233.40 g/mol) precipitate? $B a {(N O_{3})}_{2} (a q) + N a_{2} S O 4 (a q) \to B a S O_{4} (s) + 2 N a N O_{3} (a q)$ $Ba(NO_3)_2(aq) + Na_2SO4 (aq) -> BaSO_4(s) + 2NaNO_3(aq)$ ?

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Answer 1 · 2016-10-01T09:45:19

1.59g (3 s.f.)

Explanation:

In this reaction, every 1 mole of $N a_{2}$ $Na_2$ $S O_{4}$ $SO_4$ produces 1 mole of $B a S O_{4}$ $BaSO_4$ , as shown by the mole ratio in the equation. As the question states that all the $N a_{2}$ $Na_2$ $S O_{4}$ $SO_4$ solution reacts, we can assume that $B a {(N O_{3})}_{2}$ $Ba(NO_3)_2$ is in excess as it allows for all the $N a_{2}$ $Na_2$ $S O_{4}$ $SO_4$ to react. This means that the reaction is dependent on the moles of $N a_{2}$ $Na_2$ $S O_{4}$ $SO_4$ .

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

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If 45.5 mL of 0.150 M sodium sulfate solution reacts completely with aqueous barium nitrate, what is the mass of BaSO4BaSO_4 (233.40 g/mol) precipitate? Ba(NO3)2(aq)+Na2SO4(aq)→BaSO4(s)+2NaNO3(aq)Ba(NO_3)_2(aq) + Na_2SO4 (aq) -> BaSO_4(s) + 2NaNO_3(aq)?

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