Giải thích các bước giải:

\(\begin{array}{l}
5.\\
KOH + HCl \to KCl + {H_2}O\\
{n_{KOH}} = 0,2mol\\
{m_{HCl}} = \dfrac{{100 \times 36,5\% }}{{100\% }} = 36,5g\\
 \to {n_{HCl}} = 1mol\\
 \to {n_{HCl}} > {n_{KOH}} \to {n_{HCl}}dư\\
 \to {n_{HCl(pt)}} = {n_{KOH}} = 0,2mol\\
 \to {n_{HCl}}dư = 0,8mol \to {m_{HCl}}dư = 29,2g\\
{n_{KCl}} = {n_{KOH}} = 0,2mol \to {m_{KCl}} = 14,9g\\
6.\\
Ba{(OH)_2} + {H_2}S{O_4} \to BaS{O_4} + 2{H_2}O\\
{n_{Ba{{(OH)}_2}}} = 0,1mol\\
{m_{{H_2}S{O_4}}} = \dfrac{{200 \times 19,6\% }}{{100\% }} = 39,2g\\
 \to {n_{{H_2}S{O_4}}} = 0,4mol\\
 \to {n_{{H_2}S{O_4}}} > {n_{Ba{{(OH)}_2}}} \to {n_{{H_2}S{O_4}}}dư\\
 \to {n_{{\rm{BaS}}{O_4}}} = {n_{Ba{{(OH)}_2}}} = 0,1mol \to {m_{{\rm{BaS}}{O_4}}} = 23,3g
\end{array}\) 

Đáp án:

 

Giải thích các bước giải:

5 2KOH+2HCl----->2KCl+H2O

 

nKOH=0.2*1=0.2(mol)                  mHCl=100*36.5%=36.5g→nHCl=36.5/36.5=1<mol>

mKOH=0.2*56=11.2g

→nKOH=11.2/56=0.2

xét tỉ lệ

nKOH/2=0.1<nHCl/2=0.5

→nHCl dư=0.8 mol→mHCl=0.8*36.5=29.2 g

nKCl=nKOH=0.2<mol>

mKCl=0.2*74.5=14.9<g>

6,

Ba(OH)2+H2SO4----->BaSO4+2H2O

 

nBa(OH)2=0.1*1=0.1(mol)                  mH2SO4=200*19.6%=39.2g→nH2SO4=39.2/98=0.4<mol>

xét tỉ lệ

nBa(OH)2/1=0.1<nH2SO4/1=0.4

→nH2SO4 dư=0.3 mol→mH2SO4 DƯ=0.3*98=29.4 g

nBaSO4=nBa(OH)2=0.1<mol>

mBaSO4=0.1*233=23.3<g>

no COPY