Consider the following series of reactions:$$Cl2+2NaOH$$⟶$$NaCl+NaClO+H2O3NaClO$$⟶$$2NaCl+NaClO34NaClO3$$⟶$$3NaClO4+NaClHow$$ much $$Cl2$$ is needed to prepare $$122.5$$ g $$NaClO4$$ by above sequence?How much $$Cl2$$ is needed to prepare $$106.5$$ g of $$NaClO3$$ by the above sequence?The oxidation state of chlorine in compound sodium chlorate is

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Factor label method: (Mw$$ of $$NaClO4$$ $$=$$ $$122.5$$, Mw of $$NaClO3$$ $$=$$ $$106.5$$ g $$mol-1)x$$ g of $$Cl2=122.5$$ g $$NaClO4122.5$$ g $$NaClO4$$ per mol of $$NaClO34$$ mol $$NaClO33$$ mol $$NaClO43$$ mol $$NaClO1$$ mol $$NaClO31$$ mol $$Cl21$$ mol $$NaClO71.0$$ g $$Cl21$$ mol $$Cl2$$ $$=122.5122.5$$×$$43$$×$$31$$×$$11$$×$$71.01=284.0$$ gMole method: $$n(NaClO)=nCl2$$,$$nNaClO3=13n(NaClO)=13nCl2nNaClO4=34nNaClO3=34$$×$$13nCl2=14nCl2nNaClO4=122.5$$ g $$NaClO4122.5$$ of $$NaClO4/mol$$ $$NaClO4=1.0mol$$⋅$$NaClO4nCl2=4$$×$$1.0=4mol$$⋅$$Cl2Mass$$ of $$(Cl2) $$=$$ $$4$$ x $$71.0$$ $$g=284.0$$ g $$Cl2x$$ g $$Cl2=106.5$$ g $$NaClO3106.5$$⋅g $$NaClO3$$ per mol $$NaClO33$$ mol $$NaClO1$$ mol $$NaClO3$$ $$1$$ mol $$Cl21$$ mol $$NaClO71.0$$ g $$Cl21$$ mol $$Cl2$$ $$=106.5106.5$$×$$31$$×$$11$$×$$17.01=213.0$$ g $$Cl2NaClO3$$ is sodium chlorate

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