Question

Answer 1

a) \(m_{dd}=D.V=1,2.200=240\left(g\right)\)

\(C\%_{FeCl3}=\dfrac{18}{240}.100\%=7,5\%\)

b) \(n_{NaCl}=2,5.0,2=0,5\left(mol\right)\Rightarrow m_{NaCl}=0,5.58,5=29,25\left(g\right)\)

c) \(n_{NaOH}=\dfrac{32}{40}=0,8\left(mol\right)\)

\(C_{MddNaOH}=\dfrac{0,8}{0,2}=4\left(M\right)\)

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Answer 2

a, Ta có: \(m_{ddFeCl_3}=1,2.200=240\left(g\right)\)

\(\Rightarrow C\%_{FeCl_3}=\dfrac{18}{240}.100\%=7,5\%\)

b, \(n_{NaCl}=2,5.0,2=0,5\left(mol\right)\Rightarrow m_{NaCl}=0,5.58,5=29,25\left(g\right)\)

c, \(n_{NaOH}=\dfrac{32}{40}=0,8\left(mol\right)\Rightarrow C_{M_{NaOH}}=\dfrac{0,8}{0,2}=4\left(M\right)\)