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