What weight of BaCl_2 is required to react wi | vedclass

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(B) The reaction between barium chloride and sodium sulfate is given by the equation:$BaCl_2 + Na_2SO_4 \rightarrow BaSO_4 + 2NaCl$According to the La

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$45.6$

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(B) The reaction between barium chloride and sodium sulfate is given by the equation:$BaCl_2 + Na_2SO_4 \rightarrow BaSO_4 + 2NaCl$According to the Law of Conservation of Mass,the total mass of reactants must equal the total mass of products.Let the mass of $BaCl_2$ be $x \ g$.Mass of reactants $= (x + 24.4) \ g$Mass of products $= (46.6 + 23.4) \ g$Equating both sides:$x + 24.4 = 46.6 + 23.4$$x + 24.4 = 70.0$$x = 70.0 - 24.4$$x = 45.6 \ g$Therefore,the weight of $BaCl_2$ required is $45.6 \ g$.

What weight of $BaCl_2$ is required to react with $24.4 \ g$ of sodium sulfate to produce $46.6 \ g$ of barium sulfate and $23.4 \ g$ of sodium chloride (in $g$)?

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