Question #8053f

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Question #8053f

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Answer 1 · 2015-01-08T08:59:46

The answer is $3.0$ $3.0$ $g$ $"g"$ of $B a (S O_{4})$ $Ba(SO_4)$ will form in this reaction.

Your general chemical reaction is

$N a_{2} S O_{4 (a q)} + B a {(N O_{3})}_{2 (a q)} \to B a S O_{4} (s) + 2 N a N O_{3 (a q)}$ $Na_2SO_(4(aq)) + Ba(NO_3)_(2(aq)) -> BaSO_4(s) + 2NaNO_(3(aq))$

Notice that you've got a $1 : 1$ $1:1$ mole ratio between $B a {(N O_{3})}_{2}$ $Ba(NO_3)_2$ and $B a S O_{4}$ $BaSO_4$ ; this means that for every mole of $B a {(N O_{3})}_{2}$ $Ba(NO_3)_2$ used, 1 mole of solid will be produced.

The number of $B a {(N O_{3})}_{2}$ $Ba(NO_3)_2$ moles can be determined using its molarity:

$C = \frac{n}{V} \Rightarrow n_{B a {(N O_{3})}_{2}} = C \cdot V = 0.50$ $C=n/V => n_(Ba(NO_3)_2) = C * V = 0.50$ $M \cdot 25 \cdot 10^{- 3}$ $"M" * 25*10^(-3)$ $L$ $"L"$

$n_{B a {(N O_{3})}_{2}} = 0.013$ $n_(Ba(NO_3)_2) = 0.013$ $moles$ $"moles"$

This means of course that $n_{B a S O_{4}} = n_{B a {(N O_{3})}_{2}} = 0.013$ $n_(BaSO_4) = n_(Ba(NO_3)_2) = 0.013$ moles of solid will be formed. Knowing that $B a S O_{4}$ $BaSO_4$ 's molar mass is $233.3$ $233.3$ $g/mol$ $"g/mol"$ , the mass produced will be

$m_{B a S O_{4}} = n_{B a S O_{4}} \cdot 233.3 \frac{g}{m o l} = 0.013$ $m_(BaSO_4) = n_(BaSO_4) * 233.3 g/(mol) = 0.013$ $moles \cdot 233.3 \frac{g}{m o l}$ $"moles" * 233.3g/(mol)$

$m_{B a S O_{4}} = 3.0$ $m_(BaSO_4) = 3.0$ $g$ $"g"$

The reaction's complete ionic equation is

$2 N a_{(a q)}^{+} + S O_{4 (a q)}^{2 -} + B a_{(a q)}^{2 +} + 2 N O_{3 (a q)}^{-} \to B a S O_{4 (s)} + 2 N a_{(a q)}^{+} + 2 N O_{3 (a q)}^{-}$ $2Na_((aq))^(+) + SO_(4(aq))^(2-) + Ba_((aq))^(2+) + 2NO_(3(aq))^(-) -> BaSO_(4(s)) + 2Na_((aq))^(+) + 2NO_(3(aq))^(-)$

The net ionic equation is

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

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Question #8053f

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