Question

Answer 1

Bài 6:

a, \(C\%_{KCl}=\dfrac{20}{20+80}.100\%=20\%\)

b, Có: \(n_{Na_2O}=\dfrac{6,2}{62}=0,1\left(mol\right)\)

PT: \(Na_2O+H_2O\rightarrow2NaOH\)

____0,1_____________0,2 (mol)

\(\Rightarrow C\%_{NaOH}=\dfrac{0,2.40}{6,2+100}.100\%\approx7,53\%\)

c, Có: \(n_K=\dfrac{7,8}{39}=0,2\left(mol\right)\)

PT: \(2K+2H_2O\rightarrow2KOH+H_2\)

____0,2___________0,2____0,1 (mol)

⇒ m _{dd sau pư} = 7,8 + 150 - 0,1.2 = 157,6 (g)

\(\Rightarrow C\%_{KOH}=\dfrac{0,2.56}{157,6}.100\%\approx7,11\%\)

Bạn tham khảo nhé!

Answer 2

A) mdd = 20+80=100(g)
      C%=\(\dfrac{20}{100}\)x100%=20%
B) mdd =6,2+100=106,2(g)
      C%=\(\dfrac{6,2}{106,2}\)x100%=5,83%
C) mdd=7,8+150=157,8
      C%=\(\dfrac{7,8}{157,8}\)x100%=4,94(g)