PTHH: \(CuCl_2+2NaOH\rightarrow2NaCl+Cu\left(OH\right)_2\downarrow\)

            \(Cu\left(OH\right)_2\xrightarrow[]{t^o}CuO+H_2O\)

Ta có: \(n_{CuCl_2}=0,2\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}n_{Cu\left(OH\right)_2}=0,2\left(mol\right)=n_{CuO}\\n_{NaOH}=0,4\left(mol\right)=n_{NaCl}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Cu\left(OH\right)_2}=0,2\cdot98=19,6\left(g\right)\\m_{CuO}=0,2\cdot80=16\left(g\right)\\m_{NaCl}=0,4\cdot58,5=23,4\left(g\right)\\C\%_{NaOH}=\dfrac{0,4\cdot40}{200}\cdot100\%=8\%\end{matrix}\right.\)

a)

$FeCl_3 + 3NaOH \to Fe(OH)_3 + 3NaCl$

$2Fe(OH)_3 \xrightarrow{t^o} Fe_2O_3 + 3H_2O$

$n_{FeCl_3} = 0,3.2 = 0,6(mol)$

Theo PTHH : $n_{Fe_2O_3} = \dfrac{1}{2}n_{FeCl_3} = 0,3(mol)$
$\Rightarrow m_{Fe_2O_3} = 0,3.160 = 48(gam)$

b) Sau phản ứng, $V_{dd} = 0,3 + 0,3 = 0,6(lít)$

$n_{NaCl} = 3n_{FeCl_3} = 1,8(mol) \Rightarrow C_{M_{NaCl}} = \dfrac{1,8}{0,6} = 3M$

a, \(CuCl_2+2NaOH\rightarrow2NaCl+Cu\left(OH\right)_2\)

\(Cu\left(OH\right)_2\underrightarrow{t^o}CuO+H_2O\)

b, \(n_{CuCl_2}=\dfrac{27}{135}=0,2\left(mol\right)\)

Theo PT: \(n_{CuO}=n_{Cu\left(OH\right)_2}=n_{CuCl_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{CuO}=0,2.80=16\left(g\right)\)

c, \(n_{NaOH}=2n_{CuCl_2}=0,4\left(mol\right)\)

\(\Rightarrow C\%_{NaOH}=\dfrac{0,4.40}{200}.100\%=8\%\)

\(a,PTHH:3NaOH+FeCl_3\rightarrow3NaCl+Fe\left(OH\right)_3\downarrow\\ 2Fe\left(OH\right)_3\rightarrow^{t^o}Fe_2O_3+3H_2O\uparrow\\ b,n_{FeCl_3}=1,5\cdot0,2=0,3\left(mol\right)\\ \Rightarrow n_{NaOH}=3n_{FeCl_3}=0,9\left(mol\right)\\ \Rightarrow V_{dd_{NaOH}}=\dfrac{0,9}{2}=0,45\left(l\right)\)

Theo đề: \(\left\{{}\begin{matrix}X:Fe\left(OH\right)_3\\A:NaCl\\Y:Fe_2O_3\end{matrix}\right.\)

Theo PT: \(n_{NaCl}=3n_{FeCl_3}=0,9\left(mol\right)\)

\(\Rightarrow C_{M_{NaCl}}=\dfrac{0,9}{0,45+0,2}\approx1,4M\)

\(c,\) Theo PT: \(n_{Fe\left(OH\right)_3}=n_{FeCl_3}=0,3\left(mol\right);n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe\left(OH\right)_3}=0,15\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}m_X=m_{Fe\left(OH\right)_3}=0,3\cdot107=32,1\left(g\right)\\m_Y=m_{Fe_2O_3}=0,15\cdot160=24\left(g\right)\end{matrix}\right.\)

Đáp án:

m =32,4g

mddH2SO4 = 49g

Giải thích các bước giải:

a) MgCO3 + H2SO4 → MgSO4 + H2O +CO2 ↑

MgSO4 + 2NaOH → Mg(OH)2 + Na2SO4

$Mg{(OH)_2}\buildrel {to} \over

\longrightarrow MgO + {H_2}O$

b) nCO2 = 2,24 : 22,4 = 0,1mol

nMgCO3 = nCO2 = 0,1 mol

nMgO = 12:40=0,3mol

nMgSO4 = nMgO - nMgCO3 = 0,3 - 0,1 = 0,2mol

m = mMgCO3 + mMgSO4

= 0,1 .84+0,2.120=32,4g

nH2SO4 = nCO2 = 0,1 mol

mH2SO4 = 0,1.98=9,8g

mddH2SO4 = 9,8:20.100=49g

chúc bạn học tốt

a, \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

\(FeCl_3+3KOH\rightarrow3KCl+Fe\left(OH\right)_{3\downarrow}\)

\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)

Theo PT: \(n_{HCl}=6n_{Fe_2O_3}=0,6\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,6.36,5=21,9\left(g\right)\)

b, \(n_{Fe\left(OH\right)_3}=n_{FeCl_3}=2n_{Fe_2O_3}=0,2\left(mol\right)\)

\(\Rightarrow m_{Fe\left(OH\right)_3}=0,2.107=21,4\left(g\right)\)

\(n_{KOH}=3n_{FeCl_3}=0,6\left(mol\right)\)

\(\Rightarrow C_{M_{KOH}}=\dfrac{0,6}{0,2}=3\left(M\right)\)

a)\(n_{CuSO_4}=0,4.0,5=0,2\left(mol\right)\)

\(PTHH:CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)

Mol:               0,2                0,4                   0,2

⇒ \(m_{Cu\left(OH\right)_2}=0,2.98=19,6\left(g\right)\)

b)\(C_{M\left(ddNaOH\right)}=\dfrac{0,4}{0,3}=1,3\left(M\right)\)

c)\(PTHH:Cu\left(OH\right)_2\underrightarrow{t^o}CuO+H_2O\)

Mol:                 0,2                   0,2

=> m_CuO = 0,2.80 = 16 (g)

\(n_{CuSO_4}=0.2\cdot0.5=0.1\left(mol\right)\)

\(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)

\(0.1.............0.2.................0.1..........0.1\)

\(C_{M_{Na_2SO_4}}=\dfrac{0.1}{0.3+0.2}=0.2\left(M\right)\)

\(Cu\left(OH\right)_2\underrightarrow{^{^{t^0}}}CuO+H_2O\)

\(0.1.............0.1\)

\(m_{CuO}=0.1\cdot80=8\left(g\right)\)