PTHH: \(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)

Ta có: \(\left\{{}\begin{matrix}n_{Ba\left(OH\right)_2}=0,05\cdot0,5=0,025\left(mol\right)\\n_{HCl}=0,15\cdot0,1=0,015\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,025}{1}>\dfrac{0,015}{2}\) \(\Rightarrow\) Ba(OH)₂ còn dư, dd sau p/ứ có tính kiềm

\(\Rightarrow\left\{{}\begin{matrix}n_{BaCl_2}=0,0075\left(mol\right)\\n_{Ba\left(OH\right)_2\left(dư\right)}=0,0175\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}C_{M_{BaCl_2}}=\dfrac{0,0075}{0,05+0,15}=0,0375\left(M\right)\\C_{M_{Ba\left(OH\right)_2}}=\dfrac{0,0175}{0,2}=0,0875\left(M\right)\end{matrix}\right.\)

\(n_{NaOH}=\dfrac{10}{40}=0,25\left(mol\right);n_{HNO_3}=\dfrac{10}{63}\left(mol\right)\)

\(NaOH+HNO_3\rightarrow NaNO_3+H_2O\)

Lập tỉ lệ : \(0,25>\dfrac{10}{63}\)=> Sau phản ứng NaOH dư

Dung dịch sau phản ứng gồm NaOH dư và NaNO3

\(n_{NaOH\left(dư\right)}=0,25-\dfrac{10}{63}=\dfrac{23}{252}\left(mol\right)\\ \Rightarrow m_{NaOH\left(dư\right)}=\dfrac{23}{252}.40=3,65\left(g\right)\)

\(n_{NaNO_3}=n_{HNO_3}=\dfrac{10}{63}\left(mol\right)\\ \Rightarrow m_{NaNO_3}=\dfrac{10}{63}.85=13,49\left(g\right)\)

Ta có:

\(n_{NaOH}=\dfrac{10}{40}=0,25\left(mol\right)\\ n_{HNO_3}=\dfrac{10}{63}\left(mol\right)\)

\(PTHH:\)

\(NaOH+HNO_3\rightarrow NaNO_3+H_2O\)

Ta thấy:

\(n_{NaOH}>n_{HNO_3}\left(0,25>\dfrac{10}{63}\right)\) \(nên\) \(NaOH\) \(dư\)

Theo phương trình ta có:

\(n_{NaOH\left(p.ư\right)}=n_{NaNO_3}=n_{HNO_3}=\dfrac{10}{63}\left(mol\right)\)

\(\Rightarrow n_{NaOH\left(dư\right)}=0,25-\dfrac{10}{63}=\dfrac{23}{252}\left(mol\right)\)

Nên sau phản ứng:

\(m_{NaNO_3}=\dfrac{10}{63}.85=13,49\left(gam\right)\\ \)

\(m_{NaOH\left(dư\right)}=\dfrac{23}{252}.40=3,65\left(gam\right)\)

\(a.HCl+NaOH\rightarrow NaCl+H_2O\)

PỨ trung hoà 

\(b,n_{NaOH}=0,1.1=0,1mol\\ n_{NaCl}=n_{NaOH}=n_{HCl}0,1mol\\ m=m_{HCl}=0,1.36,5=3,65g\\ c,m_{NaCl}=0,1.58,5=5,85g\\ d,n_{HCl}=\dfrac{73.10}{100.36,5}=0,2mol\\ \Rightarrow\dfrac{0,1}{1}< \dfrac{0,2}{1}\Rightarrow HCl.dư\\ n_{HCl,pứ}=n_{NaOH}=0,1mol\\ m_{HCl,dư}=\left(0,2-0,1\right).36,5=3,65g\)

\(n_{Fe_2O_3}=\dfrac{1,6}{160}=0,01\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{49.6\%}{98}=0,03\left(mol\right)\)

PTHH:

\(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

0,01         0,03                0,01   

\(C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{0,01.400}{1,6+49}.100\%=7,91\left(\%\right)\)

c, axit phản ứng hết

nHNO3=0,16 mol

nNaOH=0,25 mol

NaOH + HNO3 -> NaNO3 + H2O

bđ: 0,25__0,16

pứ: 0,16<--0,16--->0,16---->0,16

sau: 0,09__0_____0,16____0,16

NaOH dư => môi trường kiềm, quỳ chuyển xanh

b. sau pứ dd gồm NaOH 0,09 mol, NaNO3 0,16 mol

khối lượng tự tinh nha

a. PTHH:

2Al + 3H₂SO₄ ---> Al₂(SO₄)₃ + 3H₂

Cu + H₂SO₄ ---x--->

b. Theo PT: \(n_{Al}=\dfrac{2}{3}.n_{H_2}=\dfrac{2}{3}.\dfrac{6,72}{22,4}=0,2\left(mol\right)\)

=> \(m_{Al}=0,2.27=5,4\left(g\right)\)

=> \(m_{Cu}=10-5,4=4,6\left(g\right)\)

c. \(\%_{m_{Al}}=\dfrac{5,4}{10}.100\%=54\%\)

\(\%_{m_{Cu}}=100\%-54\%=46\%\)

d. Theo PT: \(n_{H_2SO_4}=n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

=> \(m_{H_2SO_4}=0,3.98=29,4\left(g\right)\)

Ta có: \(C_{\%_{H_2SO_4}}=\dfrac{29,4}{m_{dd_{H_2SO_4}}}.100\%=20\%\)

=> \(m_{dd_{H_2SO_4}}=147\left(g\right)\)

nHCl=0,6 mol

FeO+2HCl-->FeCl2+ H2O

x mol               x mol

Fe2O3+6HCl-->2FeCl3+3H2O

x mol                   2x mol

72x+160x=11,6         =>x=0,05 mol

A/ C_FeCl2=0,05/0,3=1/6 M

CFeCl3=0,1/0,3=1/3 M

CHCl du=(0,6-0,4)/0,3=2/3 M

B/ 

NaOH+ HCl-->NaCl+H2O

0,2          0,2

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

0,1           0,05

3NaOH+FeCl3-->3NaCl+Fe(OH)3

0,3            0,1

nNaOH=0,6

CNaOH=0,6/1,5=0,4M

Câu 1 bạn biết làm k chỉ mình với

Gọi CT của oxit KL là M2Om 
=> %M = 2M/(2M + 16m) = 85.22% 
<=> M = 46.13m --> ko có KLoại quen thuộc (chỉ có m=2, M = 92.26 ~ Nb = 92.9) 
Tuy nhiên, ta ko cần tìm M mà vẫn tính dc (nhưng bạn vẫn nên xem lại đề nhé)
M2Om + mH2SO4 ---> M2(SO4)m + mH2O 
n(M2Om) = 10/(2M + 16m) = 10/(2*46.13m + 16m) = 10/108.26m 
--> nH2SO4 = m*10/108.26m = 10/108.26 ~ 0.0924 mol 
=> mddH2SO4 = 0.0924*98/0.1 = 90.55g

\(n_{FeCl_3}=\dfrac{48,75}{162,5}=0,3(mol)\\ 3NaOH+FeCl_3\to Fe(OH)_3\downarrow+3NaCl\\ \Rightarrow n_{Fe(OH)_3}=0,3(mol);n_{NaOH}=n_{NaCl}=0,9(mol)\\ a,m_{Fe(OH)_3}=0,3.107=32,1(g)\\ b,m_{dd_{NaOH}}=\dfrac{0,9.40}{10\%}=360(g)\\ c,C\%_{NaCl}=\dfrac{0,9.58,5}{360+48,75-32,1}.100\%=13,98\%\\ \)

\(d,2Fe(OH)_3+3H_2SO_4\to Fe_2(SO_4)_3+6H_2O\\ \Rightarrow n_{H_2SO_4}=0,45(mol)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{0,45.98}{20\%}=220,5(g)\\ \Rightarrow V_{dd_{H_2SO_4}}=\dfrac{220,5}{1,14}=193,42(ml)\)

nH2SO4= 2.0,05=0,1(mol)

PTHH: 2 NaOH + H2SO4 -> Na2SO4 + 2 H2O

a) nNaOH= 0,1.2=0,2(mol) => mNaOH=0,2.40=8(g)

=>mddNaOH= 8/20%= 40(g)

b) nNa2SO4=nH2SO4=0,1(mol)

=> mNa2SO4=142.0,1=14,2(g)