If a mass of $3.88g$ of $"barium sulfate"$ is collected from from a reaction mixture that had contained barium chloride, and sodium sulfate, $2.36g$ , what are the molar quantities of reactants and products?

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Answer 1 · 2018-02-21T18:14:44

We interrogate the stoichiometric equation...

$BaCl_2(aq) + Na_2SO_4(aq) rarr BaSO_4(s)darr+ 2NaCl(aq)$

$"Moles of sodium sulfate"=(2.36*g)/(142.04 *g*mol^-1)=0.0166*mol$ .

$"Moles of barium sulfate"=(3.88*g)/(233.38 *g*mol^-1)=0.0166*mol$ .

$"Moles of sodium chloride"=(1.94*g)/(58.44 *g*mol^-1)=0.0332*mol$ .

And so we needs an equivalent mass of $"barium chloride.."$

$0.0166*molxx208.23*g*mol^-1=3.46*g$ ...

If a mass of 3.88*g3.88*g of "barium sulfate""barium sulfate" is collected from from a reaction mixture that had contained barium chloride, and sodium sulfate, 2.36*g2.36*g, what are the molar quantities of reactants and products?