Đáp án:

Bài 9:

a, `a=m_{Fe}=25,2\ (g).`

b, `C_{M_{H_2SO_4}}=0,75\ (M).`

c, \(m_{FeSO_4.\text{7H$_2$O}}=125,1\ (g).\)

Bài 10:

a, `V=V_{H_2\ \text{(đktc)}}=1,12\ (l).`

b,

`-` `m_{BaSO_4}=5,825\ (g).`

`-` `C%_{H_2SO_4\ \text{(dư)}}=2,78%`

`-` `C%_{HCl}=1,48%`

Giải thích các bước giải:

Bài 9:

`-` `n_{H_2}=\frac{10,08}{22,4}=0,45\ (mol).`

a,

Phương trình hóa học:

`Fe + H_2SO_4 \to FeSO_4 + H_2\uparrow`

`-` Theo phương trình: `n_{Fe}=n_{H_2}=0,45\ (mol).`

`\to a=m_{Fe}=0,45\times 56=25,2\ (g).`

b,

`-` Đổi `600\ (ml)=0,6\ (l).`

Theo phương trình: `n_{H_2SO_4}=n_{H_2}=0,45\ (mol).`

`\to C_{M_{H_2SO_4}}=\frac{0,45}{0,6}=0,75\ (M).`

c,

`-` Bảo toàn nguyên tố `Fe:`

\(\to n_{FeSO_4.\text{7H$_2$O}}=n_{Fe}=0,45\ (mol).\)

\(\to m_{FeSO_4.\text{7H$_2$O}}=0,45\times 278=125,1\ (g).\)

Bài 10:

a,

`-` `n_{Fe}=\frac{4,2}{56}=0,075\ (mol).`

`-` Đổi `50\ (ml)=0,05\ (l).`

`\to n_{H_2SO_4}=0,05\times 1=0,05\ (mol).`

Phương trình hóa học:

`Fe + H_2SO_4 \to FeSO_4 + H_2\uparrow`

`-` Tỉ lệ: `n_{Fe}:n_{H_2SO_4}=\frac{0,075}{1}>\frac{0,05}{1}`

`\to Fe` dư, `H_2SO_4` hết.

`-` Theo phương trình: `n_{H_2}=n_{H_2SO_4}=0,05\ (mol).`

`\to V=V_{H_2\ \text{(đktc)}}=0,05\times 22,4=1,12\ (l).`

b,

`-` `n_{H_2SO_4}=\frac{29,4\times 20%}{98}=0,06\ (mol).`

`-` `n_{BaCl_2}=\frac{100\times 5,2%}{208}=0,025\ (mol).`

Phương trình hóa học:

`H_2SO_4 + BaCl_2 \to BaSO_4\downarrow + 2HCl`

`-` Tỉ lệ: `n_{H_2SO_4}:n_{BaCl_2}=\frac{0,06}{1}>\frac{0,025}{1}`

`\to H_2SO_4` dư, `BaCl_2` hết.

`\to n_{H_2SO_4\ \text{(dư)}}=0,06-0,025=0,035\ (mol).`

`-` Theo phương trình: `n_{BaSO_4}=n_{BaCl_2}=0,025\ (mol).`

`\to m_{BaSO_4}=0,025\times 233=5,825\ (g).`

`-` Theo phương trình: `n_{HCl}=2n_{BaCl_2}=0,05\ (mol).`

`-` `m_{\text{dd spư}}=m_{dd\ H_2SO_4}+m_{dd\ BaCl_2}-m_{BaSO_4}`

`\to m_{\text{dd spư}}=29,4+100-5,825=123,575\ (g).`

`\to C%_{H_2SO_4\ \text{(dư)}}=\frac{0,035\times 98}{123,575}\times 100%=2,78%`

`\to C%_{HCl}=\frac{0,05\times 36,5}{123,575}\times 100%=1,48%`

9)

Phản ứng xảy ra:

\(Fe + {H_2}S{O_4}\xrightarrow{{}}FeS{O_4} + {H_2}\)

Ta có:

\({n_{{H_2}}} = \frac{{10,08}}{{22,4}} = 0,45{\text{ mol = }}{{\text{n}}_{FeS{O_4}}} = {n_{{H_2}S{O_4}}} = {n_{Fe}}\)

\( \to a = {m_{Fe}} = 0,45.56 = 25,2{\text{ gam}}\)

\( \to {C_{M{\text{ }}{{\text{H}}_2}S{O_4}}} = \frac{{0,45}}{{0,6}} = 0,75M\)

Ta có:

\({n_{FeS{O_4}.7{H_2}O}} = {n_{FeS{O_4}}} = 0,45{\text{ mol}}\)

\( \to {m_{FeS{O_4}.7{H_2}O}} = 0,45.(56 + 96 + 18.7) = 125,1{\text{ gam}}\)

10)

a)

Phản ứng xảy ra:

\(Fe + {H_2}S{O_4}\xrightarrow{{}}FeS{O_4} + {H_2}\)

Ta có:

\({n_{Fe}} = \frac{{4,2}}{{56}} = 0,075{\text{ mol;}}{{\text{n}}_{{H_2}S{O_4}}} = 0,05.1 = 0,05{\text{ mol < }}{{\text{n}}_{Fe}}\)

Vậy \(Fe\) dư

\( \to {n_{{H_2}}} = {n_{{H_2}S{O_4}}} = 0,05{\text{ mol}}\)

\( \to V = {V_{{H_2}}} = 0,05.22,4 = 1,12{\text{ lít}}\)

b)

Phản ứng xảy ra:

\({H_2}S{O_4} + BaC{l_2}\xrightarrow{{}}BaS{O_4} + 2HCl\)

Ta có:

\({m_{{H_2}S{O_4}}} = 29,4.20\%  = 5,88{\text{ gam}}\)

\({m_{BaC{l_2}}} = 100.5,2\%  = 5,2{\text{ gam}}\)

\( \to {n_{{H_2}S{O_4}}} = \frac{{5,88}}{{98}} = 0,06{\text{ mol;}}{{\text{n}}_{BaC{l_2}}} = \frac{{5,2}}{{137 + 71}} = 0,025{\text{ mol < }}{{\text{n}}_{{H_2}S{O_4}}}\)

Vậy axit dư

\( \to {n_{BaS{O_4}}} = {n_{BaC{l_2}}} = 0,025{\text{ mol}}\)

\( \to {m_{BaS{O_4}}} = 0,025.(137 + 96) = 5,825{\text{ gam}}\)

BTKL:

\({m_{dd}} = {m_{dd{\text{ }}{{\text{H}}_2}S{O_4}}} + {m_{dd\;{\text{BaC}}{{\text{l}}_2}}} - {m_{BaS{O_4}}} = 29,4 + 100 - 5,825 = 123,575{\text{ gam}}\)

Ta có:

\({n_{{H_2}S{O_4}{\text{ dư}}}} = 0,06 - 0,025 = 0,035{\text{ mol;}}{{\text{n}}_{HCl}} = 0,025.2 = 0,05{\text{ mol}}\)

\( \to {m_{{H_2}S{O_4}}} = 0,035.98 = 3,43{\text{ gam;}}{{\text{m}}_{HCl}} = 0,05.36,5 = 1,825{\text{ gam}}\)

\( \to C{\% _{{H_2}S{O_4}}} = \frac{{3,43}}{{123,575}} = 2,77\% ;C{\% _{HCl}} = \frac{{1,825}}{{123,575}} = 1,47\% \)