5.11

Ba(OH)2+CO2→BaCO3+H2O

nCO2=2,24/22,4=0,1mol

→nBa(OH)2=0,1mol

CM Ba(OH)2=0,1/0,2=0,5M

mBaCO3=0,1.197=19,7g

5.12

H2O+SO3→H2SO4

nSO3=8/80=0,1mol

→nH2SO4=0,1mol

CM H2SO4=0,1/0,5=0,2M

CuO+H2SO4→CuSO4+H2O

nCuO=10/80=0,125mol

nCuO pứ=nH2SO4=0,1mol

→nCuO dư=0,125-0,1=0,025mol

mCuO=0,025.80=2g

5.13

ZnO+H2SO4→ZnSO4+H2O

nZnO=8,1/81=0,1mol

nH2SO4=0,58.0,4=0,232mol

nH2SO4 pứ=0,1mol

mH2SO4 pứ=0,1.98=9,8g

mZnSO4=0,1.161=16,1g

→nH2SO4 dư=0,232-0,1=0,132mol

CM ZnSO4=0,1/0,58=0,17M
CM H2SO4 dư=0,132/0,58=0,23M

11)

Phản ứng xảy ra:

\(Ba{(OH)_2} + C{O_2}\xrightarrow{{}}BaC{O_3} + {H_2}O\)

Ta có:

\({n_{C{O_2}}} = \frac{{2,24}}{{22,4}} = 0,1{\text{ mol}}\)

Theo phản ứng:

\({n_{Ba{{(OH)}_2}}} = {n_{BaC{O_3}}} = {n_{C{O_2}}} = 0,1{\text{ mol}}\)

Đổi 200 ml=0,2 lít

\( \to {C_{M{\text{ Ba(OH}}{{\text{)}}_2}}} = \frac{{0,1}}{{0,2}} = 0,5M\)

Kết tủa là \(BaCO_3\)

\( \to {m_{BaC{O_3}}} = 0,1.197 = 19,7{\text{ gam}}\)

2)

Phản ứng xảy ra:

\(S{O_3} + {H_2}O\xrightarrow{{}}{H_2}S{O_4}\)

Ta có:

\({n_{S{O_3}}} = \frac{8}{{32 + 16.3}} = 0,1{\text{ mol = }}{{\text{n}}_{{H_2}S{O_4}}}\)

Đổi 500ml=0,5 lít

\( \to {C_{M\;{{\text{H}}_2}S{O_4}}} = \frac{{0,1}}{{0,5}} = 0,2M\)

Cho \(CuO\) vào dung dịch

\(CuO + {H_2}S{O_4}\xrightarrow{{}}CuS{O_4} + {H_2}O\)

Ta có:

\({n_{CuO}} = \frac{{10}}{{64 + 16}} = 0,125{\text{ mol > }}{{\text{n}}_{{H_2}S{O_4}}}\)

Vậy \(CuO\) dư

\( \to {n_{CuO{\text{ dư}}}} = 0,125 - 0,1 = 0,025{\text{ mol}}\)

\( \to {m_{CuO{\text{ dư}}}} = 0,025.(64 + 16) = 2{\text{ gam}}\)

3)

Phản ứng xảy ra:

\(ZnO + {H_2}S{O_4}\xrightarrow{{}}ZnS{O_4} + {H_2}O\)

Ta có:

\({n_{ZnO}} = \frac{{8,1}}{{65 + 16}} = 0,1{\text{ mol;}}{{\text{n}}_{{H_2}S{O_4}}} = 0,58.4 = 2,32{\text{ mol > }}{{\text{n}}_{ZnO}}\)

Vậy axit dư

\( \to {n_{{H_2}S{O_4}{\text{ phản ứng}}}} = {n_{ZnS{O_4}}} = {n_{ZnO}} = 0,1{\text{ mol}}\)

\( \to {m_{{H_2}S{O_4}{\text{ phản ứng}}}} = 0,1.98 = 9,8{\text{ gam}}\)

\( \to {m_{ZnS{O_4}}} = 0,1.(65 + 96) = 16,1{\text{ gam}}\)

Ta có: 580 ml= 0,58 lít

\({n_{{H_2}S{O_4}{\text{ dư}}}} = 2,32 - 0,1 = 2,22{\text{ mol}}\)

\( \to {C_{M{\text{ ZnS}}{{\text{O}}_4}}} = \frac{{0,1}}{{0,58}} = 0,172M\)

\({C_{M\;{{\text{H}}_2}S{O_4}{\text{ dư}}}} = \frac{{2,22}}{{0,58}} = 3,83M\)