okok :)

a)

Có \(\left\{{}\begin{matrix}NTK_X=2,5.NTK_Y\\NTK_Z=5.NTK_Y\\NTK_X+NTK_Y+NTK_Z=4,25.32=136\end{matrix}\right.\)

=> NTK_Y = 16 (đvC)

=> NTK_X = 40 (đvC); NTK_Z = 80 (đvC)

X là Canxi(Ca); Y là Oxi (O); Z là Brom (Br)

b)

Xét \(\dfrac{NTK_{Br}}{NTK_{Ca}}=\dfrac{80}{40}=2\)

=> Nguyên tử Br nặng gấp 2 lần nguyên tử Ca

Bài 4.15:

a) \(n_{H_2SO_4}=1.0,02=0,02\left(mol\right)\)

PTHH: \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

                0,04<-----0,02

=> m_NaOH = 0,04.40 = 1,6 (g)

=> \(m_{dd.NaoH}=\dfrac{1,6.100}{20}=8\left(g\right)\)

b)

PTHH: \(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)

              0,04<-----0,02

=> m_KOH = 0,04.56 = 2,24 (g)

=> \(m_{dd.KOH}=\dfrac{2,24.100}{5,6}=40\left(g\right)\)

=> \(V_{dd.KOH}=\dfrac{40}{1,045}=\dfrac{8000}{209}\left(ml\right)\)

Bài 4,16

a) 

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

           0,15<--0,3<-----0,15<--0,15

=> m = 0,15.56 = 8,4 (g)

b) \(C_{M\left(dd.HCl\right)}=\dfrac{0,3}{0,05}=6M\)

c) 

\(C_{M\left(FeCl_2\right)}=\dfrac{0,15}{0,05}=3M\)

CTHH: C_aH_b

\(PTK_A=1,75.16=28\left(đvC\right)\)

Ta có: \(a=\dfrac{28.85,71\%}{12}=2\)

\(b=\dfrac{28.14,29\%}{1}=4\)

CTHH: C₂H₄

a) \(n_{CH_4}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

\(n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)

Xét tỉ lệ: \(\dfrac{0,25}{1}>\dfrac{0,3}{2}\) => CH₄ dư, O₂ hết

PTHH: \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)

            0,15<--0,3--->0,15

\(m_{CH_4\left(dư\right)}=\left(0,25-0,15\right).16=1,6\left(g\right)\)

b) \(V_{CO_2}=0,15.22,4=3,36\left(l\right)\)

\(n_{BaCl_2}=\dfrac{50.20,8\%}{208}=0,05\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{50.9,8\%}{98}=0,05\left(mol\right)\)

Khi tiếp tục thêm H₂SO₄ chỉ thấy khí thoát ra

=> Ba²⁺ đã bị kết tủa kết 

\(n_{CO_2}=\dfrac{0,0448}{22,4}=0,002\left(mol\right)\)

PTHH: \(BaCl_2+Na_2CO_3\rightarrow BaCO_3+2NaCl\)

              0,05----->0,05-------->0,05----->0,1

             \(Na_2CO_3+H_2SO_4\rightarrow Na_2SO_4+CO_2+H_2O\)

                0,002<-----0,002<-----------------0,002

\(n_{Na_2CO_3}=0,05+0,002=0,052\left(mol\right)\Rightarrow C\%_{dd.Na_2CO_3}=\dfrac{0,052.106}{100}.100\%=5,512\%\)

m_{dd sau cùng} = 50 + 100 - 0,05.197 + 50 - 0,002.44 = 190,062 (g)

Bài 3.6

a) 
\(n_{Zn}=\dfrac{26}{65}=0,4\left(mol\right)\)

PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

             0,4--->0,8------>0,4--->0,4

m_HCl = 0,8.36,5 = 29,2 (g)

\(m_{dd.HCl}=\dfrac{29,2.100}{14,6}=200\left(g\right)\)

\(V_{dd.HCl}=\dfrac{200}{1,115}=179,37\left(ml\right)\)

b) m_{dd sau pư} = 26 + 200 - 0,4.2 = 225,2 (g)

\(C\%_{ZnCl_2}=\dfrac{0,4.136}{225,2}.100\%=24,16\%\)

Bài 3.7

a)

\(n_{H_2SO_4}=1,6.0,125=0,2\left(mol\right)\)

PTHH: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

           0,2<-----0,2----->0,2

=> m_Zn = 0,2.65 = 13 (g) 

=> m_Cu = 21 - 13 = 8 (g)

\(\left\{{}\begin{matrix}\%Cu=\dfrac{8}{21}.100\%=38,1\%\\\%Zn=\dfrac{13}{21}.100\%=61,9\%\end{matrix}\right.\)

b)

\(C_{M\left(ZnSO_4\right)}=\dfrac{0,2}{0,125}=1,6M\)

ultr :))))

a)

Gọi \(\left\{{}\begin{matrix}n_{HCl}=3a\left(mol\right)\\n_{H_2SO_4}=a\left(mol\right)\end{matrix}\right.\)

m_NaOH = 20.0,5 = 10 (g)

\(n_{NaOH}=\dfrac{10}{40}=0,25\left(mol\right)\)

PTHH: \(NaOH+HCl\rightarrow NaCl+H_2O\)

            \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

Theo PTHH: n_NaOH = 3a + 2a = 0,25 (mol)

=> a = 0,05

\(\left\{{}\begin{matrix}C_{M\left(HCl\right)}=\dfrac{0,15}{0,1}=1,5M\\C_{M\left(H_2SO_4\right)}=\dfrac{0,05}{0,1}=0,5M\end{matrix}\right.\)

b) \(\left\{{}\begin{matrix}n_{HCl}=1,5.0,2=0,3\left(mol\right)\\n_{H_2SO_4}=0,5.0,2=0,1\left(mol\right)\end{matrix}\right.\)

\(\left\{{}\begin{matrix}n_{NaOH}=0,001.V.0,2=0,0002V\left(mol\right)\\n_{Ba\left(OH\right)_2}=0,001.V.0,1=0,0001V\left(mol\right)\end{matrix}\right.\)

Gọi công thức chung của axit là HA

n_HA = 0,3 + 0,1.2 = 0,5 (mol)

Gọi công thức chung của bazo là BOH

n_BOH = 0,0002V + 2.0,0001V = 0,0004V (mol)

PTHH: \(BOH+HA\rightarrow BA+H_2O\)

Theo PTHH: n_BOH = 0,0004V = 0,5

=> V = 1250 (ml)

a) \(n_{HCl\left(bđ\right)}=0,3.2=0,6\left(mol\right)\Rightarrow n_{HCl\left(pư\right)}=0,6:125\%=0,48\left(mol\right)\)

PTHH: \(2M+2nHCl\rightarrow2MCl_n+nH_2\)

           \(\dfrac{0,48}{n}\)<--0,48

=> \(M_M=\dfrac{4,32}{\dfrac{0,48}{n}}=9n\left(g/mol\right)\)

Xét n = 3 thỏa mãn => M_M = 27 (g/mol)

=> M là Al

b) 
PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

                        0,48---0,16

=> \(\left\{{}\begin{matrix}C_{M\left(AlCl_3\right)}=\dfrac{0,16}{0,3}=0,53M\\C_{M\left(HCl.dư\right)}=\dfrac{0,6-0,48}{0,3}=0,4M\end{matrix}\right.\)