V_FeCl2 = 100 ml = 0,1 (l)

=> n_FeCl2 = 0,1 . 1 = 0,1 mol

FeCl₂ + 2NaOH->2 NaCl + Fe(OH)₂ \(\downarrow\)

0,1----------------------------->0,1 mol

Fe(OH)₂ \(^{to}\rightarrow\) FeO + H₂O

0,1---------->0,1mol

FeO + CO \(^{to}\rightarrow\) Fe + CO₂ \(\uparrow\)

0,1--------------->0,1mol

=> m_Fe = 0,1 . 56 = 5,6 g

vậy khối lượng KL màu trắng bạc (Fe) là 5,6 g

a, \(MgCl_2+2NaOH\rightarrow Mg\left(OH\right)_2+2NaCl\)

\(Mg\left(OH\right)_2\underrightarrow{t^o}MgO+H_2O\)

b, \(n_{MgCl_2}=0,2.0,25=0,05\left(mol\right)\)

 Theo PT: \(n_{MgO}=n_{Mg\left(OH\right)_2}=n_{MgCl_2}=0,05\left(mol\right)\)

\(\Rightarrow m_{MgO}=0,05.40=2\left(g\right)\)

c, \(n_{NaOH}=2n_{MgCl_2}=0,1\left(mol\right)\)

\(\Rightarrow m_{ddNaOH}=\dfrac{0,1.40}{15\%}=\dfrac{80}{3}\left(g\right)\)

Đáp án B

Fe³⁺+ 3OH^- → Fe(OH)₃

0,1 mol           0,1 mol

Al³⁺+ 3OH^- → Al(OH)₃

Zn²⁺+ 2OH^- → Zn(OH)₂

Do Al(OH)₃ và Zn(OH)₂ là hidroxit lưỡng tính nên bị tan trong NaOH dư

Kết tủa thu được có Fe(OH)₃

2Fe(OH)₃  → Fe₂O₃+ 3H₂O

0,1               0,05 mol

m_Fe2O3 = 0,05.160 = 8 gam

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

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

b, \(m_{CuSO_4}=250.16\%=40\left(g\right)\Rightarrow n_{CuSO_4}=\dfrac{40}{160}=0,25\left(mol\right)\)

Theo PT: \(n_{CuO}=n_{Cu\left(OH\right)_2}=n_{CuSO_4}=0,25\left(mol\right)\)

\(\Rightarrow a=m_{CuO}=0,25.80=20\left(g\right)\)

c, Ta có: m _{dd sau pư} = m _{dd NaOH} + m _{dd CuSO4} - m_Cu(OH)2 = 200 + 250 - 0,25.98 = 425,5 (g)

\(a)2NaOH+CuSO_4\rightarrow Cu\left(OH\right)_2+Na_2SO_4\\ Cu\left(OH\right)_2\xrightarrow[]{t^0}CuO+H_2O\\ b)n_{CuSO_4}=\dfrac{250.16}{100.160}=0,25mol\\ n_{CuSO_4}=n_{Cu\left(OH\right)_2}=n_{CuO}=0,25mol\\ a=m_{CuO}=0,25.80=20g\\ c)m_{dd}=200+250-0,25.98=425,5g\)

\(n_{AlCl_3}=0.2\cdot1=0.2\left(mol\right)\)

\(n_{NaOH}=0.5V\left(mol\right)\)

\(n_{Al_2O_3}=\dfrac{5.1}{102}=0.05\left(mol\right)\)

\(2Al\left(OH\right)_3\underrightarrow{^{^{t^0}}}Al_2O_3+3H_2O\)

\(0.1...............0.05\)

TH1 : Al(OH)₃ không bị hòa tan.

\(AlCl_3+3NaOH\rightarrow Al\left(OH\right)_3+3NaCl\)

\(0.1...........0.3................0.1\)

\(\Leftrightarrow V=\dfrac{0.3}{0.5}=0.6\left(l\right)\)

TH2 : Al(OH)₃ bị hòa tan một phần 
\(AlCl_3+3NaOH\rightarrow Al\left(OH\right)_3+3NaCl\)

\(0.2...........0.6................0.2\)

\(NaOH+Al\left(OH\right)_3\rightarrow NaAlO_2+2H_2O\)

\(0.5V-0.6...0.5V-0.6\)

\(n_{Al\left(OH\right)_3}=0.2+0.5V-0.6=0.1\left(mol\right)\)

\(\Rightarrow V=1\left(l\right)\)

Đặt a, b, c là số mol Al203, CuO , FeO mA = 102a + 80b + 72c = 20,3 m rắn = 102a + 64b + 56c = 17,1 nHCl = 6a + 2b + 2c = 0,7 -> a = 0,05 ; b = c = 0,1 Trong B chứa nCO2 = b + c = 0,2 nCaCO3 = 0,1 -> nCa(HCO3)2 = 0,05 -> nCa(OH)2 = 0,15 -> Vdd =105 ml

Đáp án A