Question

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Answer 1

3) \(n_{NaOH}=0,12.1,5=0,18\left(mol\right)\)

PTHH: NaOH + HCl ---> NaCl + H₂O

            0,18----->0,18--->0,18

=> \(V_{ddHCl}=\dfrac{0,18}{2}=0,09\left(l\right)\)

=> \(V_{dd.sau.pư}=0,09+0,12=0,21\left(l\right)\)

=> \(C_{M\left(NaCl\right)}=\dfrac{0,18}{0,21}=\dfrac{6}{7}M\)

PTHH: 2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O

            0,18------>0,09

=> \(m_{ddH_2SO_4}=\dfrac{0,09.98}{20\%}=44,1\left(g\right)\)

=> \(V_{ddH_2SO_4}=\dfrac{44,1}{1,25}=35,28\left(ml\right)\)