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!
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)
