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giúp á ássssssssssssssssssssssssssss

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    câu `6:`

    `n_{Cu(NO_3)_2}=0,1.2=0,2(mol)`

    `n_{NaOH}=0,1.3=0,3(mol)`

    `PTHH:Cu(NO_3)_2+2NaOH->Cu(OH)_2+2NaNO_3`

    ta có : `{0,2}/1 >{0,3}/2=>Cu(NO_3)_2` dư

    `a.n_{Cu(OH)_2}=1/2 n_{NaOH}=1/2 .0,3=0,15(mol)`

    `m_{Cu(OH)_2}=0,15.98=14,7(g)`

    `b.“dd X:Cu(NO_3)_2 dư, NaNO_3`

    `n_{Cu(NO_3)_2 pư }=1/2 n_{NaOH}=0,15(mol)`

    `=>n_{Cu(NO_3)_2} dư=0,2-0,15=0,05(mol)`

    `C_M Cu(NO_3)_2={0,05}/{0,2}=0,25(M)`

    `n_{NaNO_3}=n_{NaOH}=0,3(mol)`

    `=>C_M NaNO_3 ={0,3}/{0,2}=1,5(M)`

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