How many grams of solid BaSO4 will form when Na2SO4 reacts with 25 mL of 0.50 M Ba(NO3)2? Ba(NO3)2 + Na2SO4 BaSO4 + NaNO3

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How many grams of solid BaSO4 will form when Na2SO4 reacts with 25 mL of 0.50 M Ba(NO3)2? Ba(NO3)2 + Na2SO4 BaSO4 + NaNO3

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Answer 1 · 2014-12-14T14:10:09

The answer is $2.79 g$ $2.79g$ .

Starting from the balanced chemical equation

$B a {(N O_{3})}_{2 (a q)} + N a_{2} S O_{4 (a q)} \to B a S O_{4 (s)} + 2 N a N O_{3 (a q)}$ $Ba(NO_3)_(2(aq)) + Na_2SO_(4(aq)) -> BaSO_(4(s)) + 2NaNO_(3(aq))$

Taking into consideration the solubility rules (more here: http://www.chem.sc.edu/faculty/morgan/resources/solubility/), we can see that $B a {(N O_{3})}_{2}$ $Ba(NO_3)_2$ , $N a_{2} S O_{4}$ $Na_2SO_4$ , and $N a N O_{3}$ $NaNO_3$ will dissociate into their respective ions, which will lead to the reaction's net ionic equation

$B a_{(a q)}^{2 +} + S O_{4 (a q)}^{2 -} \to B a S O_{4 (s)}$ $Ba_((aq))^(2+) + SO_(4(aq))^(2-) -> BaSO_(4(s))$

Since sulfate compounds formed with $B a^{2 +}$ $Ba^(2+)$ cations are insoluble in water (only slightly soluble, $B a S O_{4}$ $BaSO_4$ 's $K_{s p}$ $K_(sp)$ being equal to $1.1 \cdot 10^{- 10}$ $1.1 * 10^(-10)$ ), this double-replacement reaction forms a precipitate, $B a S O_{4}$ $BaSO_4$ .

We know from the balanced chemical equation that the mole-to-mole ratio of $B a {(N O_{3})}_{2}$ $Ba(NO_3)_2$ and $B a S O_{4}$ $BaSO_4$ is 1:1; that is, for every mole of barium nitrate used, one mole of barium sulfate is produced.

The number of barium nitrate moles can be determined from its molarity, $C = \frac{n}{V}$ $C = n/V$

$n_{B a {(N O_{3})}_{2}} = C \cdot V = 0.500 M \cdot 25.0 \cdot 10^{- 3} L = 0.012$ $n_(Ba(NO_3)_2) = C * V = 0.500 M * 25.0 * 10^(-3) L = 0.012$

Knowing barium sulfate's molar mass ( $233.3 \frac{g}{m o l}$ $233.3 g/(mol)$ ), and the number of moles produced, we get

$m_{B a S O_{4}} = n_{B a S O_{4}} \cdot m o l a r m a s s = 0.012 m o l e s \cdot 233.3 \frac{g}{m o l e} = 2.79 g$ $m_(BaSO_4) = n_(BaSO_4) * molarmass = 0.012 mol es * 233.3g/(mol e) = 2.79g$

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How many grams of solid BaSO4 will form when Na2SO4 reacts with 25 mL of 0.50 M Ba(NO3)2? Ba(NO3)2 + Na2SO4 BaSO4 + NaNO3

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