Đề thiếu em ạ!

Sửa đề: 10,23g CuO và PbO

Bảo toàn Cacbon: \(n_{CO_2}=n_{CaCO_3}=\dfrac{11}{100}=0,11\left(mol\right)\)

PTHH: \(CuO+CO\xrightarrow[]{t^o}Cu+CO_2\)

                 a_________________a    (mol)

            \(PbO+CO\xrightarrow[]{t^o}Pb+CO_2\)

                 b_________________b     (mol)

Ta lập HPT: \(\left\{{}\begin{matrix}80a+223b=10,23\\a+b=0,11\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,01\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{CuO}=\dfrac{0,1\cdot80}{10,23}\cdot100\%\approx78,2\%\\\%m_{PbO}=21,8\%\end{matrix}\right.\)

a) \(PbO+CO-^{t^o}\rightarrow Pb+CO_2\)

\(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)

b) \(n_{kt}=n_{CO_2}=0,11\left(mol\right)\)

\(n_{PbO}=n_{CO_2}=0,11\left(mol\right)\)

=> \(\%m_{PbO}=\dfrac{0,11.223}{10,23}.100=239,78\)%

Đề sai :D

a) Đặt : nAl=a(mol); nFe=b(mol)

PTHH: 2Al + 6 HCl -> 2 AlCl3 + 3 H2 (1)

a_________3a______a______1,5a(mol)

Fe +2  HCl -> FeCl2 +  H2 (2)

b_____2b_____b____b(mol)

Ta lập hpt:

\(\left\{{}\begin{matrix}27a+56b=8,3\\1,5a+b=\dfrac{5,6}{22,4}=0,25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\)

b) nHCl(p.ứ)=3a+2b=0,5(mol) => nHCl(lấy)=0,55(mol)

=>mHCl(lấy)=0,55.36,5=20,075(g) 

=>C%ddHCl(lấy)=(20,075/100).100=20,075%

c) mddsau= mhh + mddHCl - mH2= 8,3+ 100 - 0,25.2= 107,8(g)

mAlCl3=0,1.133,5=13,35(g); mFeCl2= 0,1.127=12,7(g)

=>C%ddAlCl3= (13,35/107,8).100=12,384%

C%ddFeCl2=(12,7/107,8).100=11,781%

d) AlCl3 + 3 NaOH -> Al(OH)3 + 3 NaCl

0,1_____0,3______0,1______0,3(mol)

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

0,1____0,2_______0,1______0,2(mol)

2 Al(OH)3 -to-> Al2O3 + 3 H2O

0,1________0,05(mol)

Fe(OH)2 -to-> FeO + H2O

0,1_______0,1(mol)

m=m(ket tua)= mFe(OH)2+ mAl(OH)3= 0,1. 90+ 0,1.78= 16,8(g)

a=m(rắn)= mFeO + mAl2O3= 0,1.72+ 0,05.102= 12,3(g)

Chúc em học tốt! 

k = (2.2 + 2 - 4)/2 = 1 ⇒ X là este no, đơn chức.

M_Z  = 2.23 = 46(đvC) mà Z là ancol nên Z là C₂H₅OH

⇒ Sai đề

Bảo toàn khối lượng: \(m_{FeCl_3}=m_{Fe\left(OH\right)_3}+m_{NaCl}-m_{NaOH}=10\left(g\right)\)

\(FeCl_3 + 3NaOH \to Fe(OH)_3 + 3NaCl\\ m_{FeCl_3} + m_{NaOH} = m_{Fe(OH)_3} + m_{NaCl}\\ \Rightarrow m_{FeCl_3} = 17 + 8 - 15 = 10(gam)\)

\(n_{H_2}=\dfrac{5.6}{22.4}=0.25\left(mol\right)\)

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

\(.........0.5............0.25\)

\(m_{HCl}=0.5\cdot36.5=18.25\left(g\right)\)

\(n_{CuO}=\dfrac{12}{80}=0.15\left(mol\right)\)

\(CuO+H_2\underrightarrow{^{t^0}}Cu+H_2O\)

\(1...........1\)

\(0.15.........0.25\)

\(LTL:\dfrac{0.15}{1}< \dfrac{0.25}{1}\Rightarrow H_2dư\)

\(n_{Cu}=n_{CuO}=0.15\left(mol\right)\)

\(m_{Cu}=0.15\cdot64=9.6\left(g\right)\)

a) Zn + 2HCl $\to$ ZnCl2 + H2

b) n H2 = 5,6/22,4 = 0,25(mol)

Theo PTHH :

n HCl = 2n H2 = 0,5(mol)

m HCl = 0,5.36,5 = 18,25(gam)

c) CuO + H2 $\xrightarrow{t^o}$ Cu + H2O

Ta có :

n CuO = 12/80 = 0,15 < n H2 = 0,25 => H2 dư

Theo PTHH :

n Cu = n CuO = 0,15 mol

=> m Cu = 0,15.64 = 9,6 gam

a,\(n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\)

PTHH: Fe + 2HCl → FeCl₂ + H₂

Mol:     0,4                   0,4      0,4

\(V_{H_2}=0,4.22,4=8,96\left(l\right)\)

b,\(m_{FeCl_2}=0,4.127=50,8\left(g\right)\)

PTHH: \(2KNO_3\xrightarrow[]{t^o}2KNO_2+O_2\uparrow\)

Theo đề bài: \(m_{giảm}=m_{O_2}=2,4\left(g\right)\) \(\Rightarrow n_{O_2}=\dfrac{2,4}{32}=0,075\left(mol\right)\)

\(\Rightarrow n_{KNO_3}=0,15\left(mol\right)\) \(\Rightarrow m_{KNO_3}=0,15\cdot101=15,15\left(g\right)\)