nFe2O3= 4,8/160= 0,03(mol)

nH2SO4= (300.9,8%)/98= 0,3(mol)

PTHH: Fe2O3 + 3 H2SO4 -> Fe2(SO4)3 + 3 H2O

Ta có: 0,03/1 < 0,3/3

=> H2SO4 dư, Fe2O3 hết.

- Chất tan trong dd thu được có Fe2(SO4)3 và H2SO4(dư)

nFe2(SO4)3= nFe2O3=0,03(mol)

=>mFe2(SO4)3= 400.0,03= 12(g)

nH2SO4(dư)= 0,3 - 0,03.3= 0,21(mol)

=>mH2SO4(dư)= 0,21. 98=20,58(g)

mddsau= mFe2O3+ mddH2SO4= 4,8+300=304,8(g)

=>C%ddFe2(SO4)3= (12/304,8).100=3,937%

C%ddH2SO4(dư)= (20,58/304,8).100=6,752%

Bài 14 : 

\(a) n_{CuO} = \dfrac{8}{80} = 0,1(mol)\\ CuO + 2HCl \to CuCl_2 + H_2O\\ n_{HCl} = 2n_{CuO} = 0,2(mol)\\ m_{dd\ HCl} = \dfrac{0,2.36,5}{7,3\%} = 100(gam)\\ b) \text{Chất tan : } CuCl_2\\ n_{CuCl_2} = n_{CuO} = 0,1(mol)\\ m_{CuCl_2} = 0,1.135 = 13,5(gam)\)

Bài 15 : 

\(a) n_{Fe_2O_3} =\dfrac{4,8}{160} = 0,03(mol)\\ Fe_2O_3 + 3H_2SO_4 \to Fe_2(SO_4)_3 + 3H_2O\\ n_{H_2SO_4} = 3n_{Fe_2O_3} = 0,09(mol)\\ m_{dd\ H_2SO_4} = \dfrac{0,09.98}{9,8\%} = 90(gam)\\ b) \text{Chất tan : } Fe_2(SO_4)_3\\ n_{Fe_2(SO_4)_3} = n_{Fe_2O_3} = 0,03(mol)\\ m_{Fe_2(SO_4)_3} = 0,03.400 = 12(gam)\)

Bài 16 :

\(a) Fe + 2HCl \to FeCl_2 + H_2\\ n_{H_2} = n_{Fe} = \dfrac{11,2}{56} = 0,2(mol)\\ V_{H_2} = 0,2.22,4 = 4,48(lít)\\ b) n_{HCl} = 2n_{Fe} = 0,4(mol)\\ C\%_{HCl} = \dfrac{0,4.36,5}{200}.100\% = 7,3\%\)

bài 1

\(n_{MgO}=\dfrac{m}{M}=\dfrac{8}{40}=0,2\left(mol\right)\)

\(m_{H_2SO_4}=\dfrac{C\%.m_{dd}}{100}=\dfrac{10.147}{100}=14,7\left(g\right)\)

\(n_{H_2SO_4}=\dfrac{m}{M}=\dfrac{14,7}{98}=0,15\left(mol\right)\)

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

TPƯ:   0,2          0,15

PƯ:     0,15         0,15       0,15           0,15

SPƯ:   0,05          0          0,15            0,15

a) \(m_{MgSO_4}=n.M=0,15.120=18\left(g\right)\)

b) theo định luật bảo toàn khối lượng

\(m_{ddspu}=m_{MgO}+m_{ddH_2SO_4}\)=8+147=155(g)

\(C\%_{MgO}=\dfrac{m_{ct}}{m_{dd}}.100=\dfrac{8}{155}.100=5,2\%\)

\(C\%_{MgSO_4}=\dfrac{18}{155}.100=11,6\%\)

bài 2

\(n_{ZnO}=\dfrac{m}{M}=\dfrac{20,25}{81}=0,25\left(mol\right)\)

\(m_{HCl}=\dfrac{C\%.m_{dd}}{100}=\dfrac{20.73}{100}14,6\left(g\right)\)

\(n_{HCl}=\dfrac{m}{M}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\)

PTHH:\(ZnO+2HCl\rightarrow ZnCl_2+H_2O\)

TPU:    0,25       0,4

PU:        0,2        0,4        0,2          0,2

SPU:      0,05      0           0,2           0,2

a)\(m_{ZnCl_2}=n.M=0,2.136=27,2\left(g\right)\)

b)theo định luật bảo toàn khối lượng 

\(m_{ddspu}=m_{ZnO}+m_{ddHCl}\)=20,25+73=93,25(g)

\(m_{ZnOdu}=n.M=0,05.81=4,05\left(g\right)\)

\(C\%_{ZnOdu}=\dfrac{4,05.100}{93,25}=4,3\%\)

\(C\%_{ZnCl_2}=\dfrac{27,2.100}{93,25}=29,2\%\)

\(n_{Na_2SO_3}=\dfrac{25,2}{126}=0,2(mol)\\ n_{HCl}=\dfrac{250.7,3\%}{100\%.36,5}=0,5(mol)\\ PTHH:Na_2SO_3+2HCl\to 2NaCl+H_2O+SO_2\uparrow\)

Vì \(\dfrac{n_{Na_2SO_3}}{1}<\dfrac{n_{HCl}}{2}\) nên \(HCl\) dư

\(a,n_{SO_2}=n_{Na_2SO_3}=0,2(mol)\\ \Rightarrow V_{SO_2}=0,2.22,4=4,48(l)\)

\(b,\) Chất tan trong dd sau phản ứng là \(NaCl\)

\(c,n_{NaCl}=2n_{Na_2SO_3}=0,4(mol)\\ \Rightarrow m_{NaCl}=0,4.58,5=23,4(g)\)

Câu trên mình sai nha

\(n_{Na_2SO_3}=\dfrac{25,2}{126}=0,2(mol)\\ n_{HCl}=\dfrac{250.7,3\%}{100\%.36,5}=0,5(mol)\\ PTHH:Na_2SO_3+2HCl\to 2NaCl+H_2O+SO_2\uparrow\)

Vì \(\dfrac{n_{Na_2SO_3}}{1}<\dfrac{n_{HCl}}{2}\) nên \(HCl\) dư

\(a,n_{SO_2}=n_{Na_2SO_3}=0,2(mol)\\ \Rightarrow V_{SO_2}=0,2.22,4=4,48(l)\)

\(b,\) Chất tan trong dd sau phản ứng gồm \(HCl\) dư và \(NaCl\)

\(c,n_{HCl(dư)}=0,5-0,2.2=0,1(mol);n_{NaCl}=0,4(mol)\\ \Rightarrow m_{\text{dd sau p/ứ}}=m_{HCl(dư)}+m_{NaCl}=0,1.36,5+0,4.58,5=27,05(g)\)

Na2SO3+2HCl->2NaCl+H2O+SO2

0,201-------0,402------0,402---------------0.201

n Na2SO3=0,201 mol

m HCl=18,25 g

->n HCl=0,5 mol

=>HCl dư

=>VSO2=0,201.22,4=4,5024l

b)

mNaCl=\(\dfrac{0,402.58,5}{25,4+250-0,201.64}\).100=8,95%

m Hcl dư=\(\dfrac{0,098.36,5}{25,4+250-0,201.64}.100=1,36\%\)

nAl2O3= 51/102= 0,5(mol)

mHCl= 3,65%.200= 7,3(g) -> nHCl=7,3/36,5= 0,2(mol)

PTHH: Al2O3 + 6 HCl ->2 AlCl3 + 3 H2O

Ta có: 0,2/6 < 0,5/1

=> Al2O3 dư và HCl hết => Tính theo nHCl

=> nAl2O3(p.ứ)= 1/6. 0,2= 1/30 (mol) => mAl2O3=102.1/30=3,4(g)

nAlCl3= 2/6. 0,2=1/15(mol) => mAlCl3=133,5. 1/15= 8,9(g)

mddAlCl3= mAl2O3(p.ứ) + mddHCl = 3,4+200=203,4(g)

=>C%ddAlCl3= (8,9/203,4).100= 4,376%

D

D

D

\(n_{MgO}=\dfrac{4}{40}=0,1\left(mol\right)\)

Pt : \(MgO+2HCl\rightarrow MgCl_2+H_2O|\)

         1            2              1             1

        0,1       0,2             0,1

a) \(n_{HCl}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)

⇒ \(m_{HCl}=0,2.36,5=7,3\left(g\right)\)

\(C_{HCl}=\dfrac{7,3.100}{300}=2,43\)⁰/₀

b) \(n_{MgCl2}=\dfrac{0,2.1}{2}=0,1\left(mol\right)\)

⇒ \(m_{MgCl2}=0,1.95=9,5\left(g\right)\)

\(m_{ddspu}=4+300=304\left(g\right)\)

\(C_{MgCl2}=\dfrac{9,5.100}{304}=3,125\)⁰/₀

 Chúc bạn học tốt

\(a,PTHH:2Na+2H_2O\rightarrow2NaOH+H_2\\ n_{Na}=\dfrac{2,3}{23}=0,1mol\\ n_{NaOH}=0,1.2=0,2mol\\ C_{M_{NaOH}}=\dfrac{0,2}{0,5}=0,4M\\ b,oxit.kl:RO\\ n_{RO}=\dfrac{2,4}{R+16}mol\\ n_{HCl}=\dfrac{30.7,3}{100.36,5}=0,06mol\\ RO+2HCl\rightarrow RCl_2+H_2O\\ \Rightarrow\dfrac{2,4}{R+16}=0,06:2\\ \Leftrightarrow R=64,Cu\)