Ta có: \(n_{MgO}=\dfrac{7}{40}=0,175\left(mol\right)\)

\(n_{H_2SO_4}=0,6.1=0,6\left(mol\right)\)

PT: \(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)

Xét tỉ lệ: \(\dfrac{0,175}{1}< \dfrac{0,6}{1}\), ta được H₂SO₄ dư.

Theo PT: \(n_{MgSO_4}=n_{H_2SO_4\left(pư\right)}=n_{MgO}=0,175\left(mol\right)\)

\(\Rightarrow n_{H_2SO_4\left(dư\right)}=0,6-0,175=0,425\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C_{M_{MgSO_4}}=\dfrac{0,175}{0,6}=\dfrac{7}{24}\left(M\right)\\C_{M_{H_2SO_4}}=\dfrac{0,425}{0,6}=\dfrac{17}{24}\left(M\right)\end{matrix}\right.\)

a)Gọi A là kim loại cần tìm.

\(n_{H_2}=\dfrac{1,12}{22,4}=0,05mol\)

Ta có PTHH:

\(2A+2H_2O\rightarrow2AOH+H_2\uparrow\)

0,1-----------------0,1--------0,05-----(mol);

Vậy \(M_A=\dfrac{2,3}{0,1}=23\)(g/mol) => A là Na

b) Ta có: \(m_{dd}=2,3+57,8-0,05\cdot2=60g\)

Từ đó suy ra:\(\%C_{NaOH}=\dfrac{0,1\cdot40}{60}\cdot100\%=6,67\%\)

PT: \(2Al+3Cl_2\underrightarrow{t^o}2AlCl_3\)

\(AlCl_3+3AgNO_3\rightarrow Al\left(NO_3\right)_3+3AgCl\)

Ta có: \(n_{AgNO_3}=0,1.0,6=0,06\left(mol\right)\)

Theo PT: \(n_{Al}=n_{AlCl_3}=\dfrac{1}{3}n_{AgNO_3}=0,02\left(mol\right)\)

⇒ m = m_Al = 0,02.27 = 0,54 (g)

\(C_{M_{AlCl_3}}=\dfrac{0,02}{0,15}=\dfrac{2}{15}\left(M\right)\)

M + H2O -> MOH + 1/2 H2

nH2= 0,075(mol) -> nM=0,15(mol)

=>M(M)= mM/nM= 5,85/0,15=39(g/mol)

=> M(I) cần tìm là Kali (K=39)

K + H2O -> KOH + 1/2 H2

nKOH= nK=0,15(mol) => mKOH=0,15.56=8,4(g)

mddKOH=mK + mH2O -  mH2= 5,85+194,3 - 0,075.2= 200(g)

=>C%ddKOH= (8,4/200).100=4,2%

=>CHỌN B

a.\(n_{NaOH}=\dfrac{8}{40}=0,2mol\)

\(V_{dd}=\dfrac{120}{1,2}=100ml=0,1l\)

\(C_{M_{NaOH}}=\dfrac{0,2}{0,1}=2M\)

b.\(n_{NaOH}=\dfrac{21,6}{40}=0,54mol\)

\(V_{dd}=\dfrac{180}{1,2}=150ml=0,15l\)

\(C_{M_{NaOH}}=\dfrac{0,54}{0,15}=3,6M\)

Đáp án B

Độ tan : \(S = \dfrac{20,7}{100}.100 = 20,7(gam)\)

Công thức tính số tan : S = \(\dfrac{m_{chất\ tan}}{m_{dung\ môi}}\)

S(CuSO4,200C)= (20,7.100)/100=20,7(g)

=> Chọn B

S(CuSO4,200C)= (20,7.100)/100=20,7(g