Question

Cho 18,8g K2O vào 400g đ NaOH 6%. Tính C% dung dịch mới

Answer 1

PTHH : K₂O + H₂O--> 2 KOH;

Theo pt : n_KOH= 2 n_K2O= 2* 18,8/94= 0,4 mol;

m_{dd mới} = 18,8 + 400 = 418,8 (g);

m_{chất tan trong dd mới} = 0,4* 56 + 400*6%=46,4g;

C% = 46,4/418,8 =11,08%

Answer 2

nK2O= 18.8/94=0.2 mol

mNaOH (bđ)= 400*6/100=24g

K2O + H2O --> 2KOH

0.2____________0.4

mdd sau phản ứng= mK2O + mddNaOH= 18.8 + 400=418.8 g

m chất tan sau phản ứng= mKOH + mNaOH= 0.4*56 + 24=46.4g

C% = 46.4/418.8*100%= 11.08%

Answer 3

K₂O + H₂O → 2KOH

\(n_{K_2O}=\frac{18,8}{94}=0,2\left(mol\right)\)

Theo PT: \(n_{KOH}=2n_{K_2O}=2\times0,2=0,4\left(mol\right)\)

\(\Rightarrow m_{KOH}=0,4\times56=22,4\left(g\right)\)

\(m_{NaOH}=400\times6\%=24\left(g\right)\)

\(\Rightarrow\Sigma m_{chấttan}mới=m_{KOH}+m_{NaOH}=22,4+24=46,4\left(g\right)\)

\(\Sigma m_{dungdịch}mới=m_{K_2O}+m_{ddNaOH}=18,8+400=418,8\left(g\right)\)

\(\Rightarrow C\%_{dungdịch}mới=\frac{46,4}{418,8}\times100\%\approx11,08\%\)