Bài 42: Nồng độ dung dịch

Câu 4:

Sau p/ứ nung nóng, ta thấy CO còn dư, oxit p/ứ hết

a) Ta có: \(\left\{{}\begin{matrix}n_{hh\left(khí\right)}=\dfrac{2,688}{22,4}=0,12\left(mol\right)\\n_{O\left(oxit\right)}=\dfrac{m_{oxit}-m_{KL}}{16}=\dfrac{6,22-4,94}{16}=0,08\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}n_{O_2}=0,04\left(mol\right)=n_{CO_2}\\n_{CO}=0,12-0,04=0,08\left(mol\right)\end{matrix}\right.\) \(\Rightarrow d_{D/H_2}=\dfrac{0,04\cdot44+0,08\cdot28}{2}=2\)

b) PTHH: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)

Ta có: \(n_{H_2}=\dfrac{1,568}{22,4}=0,07\left(mol\right)=\Sigma n_{Fe}\)

PTHH: \(Fe_3O_4+4CO\xrightarrow[]{t^o}3Fe+4CO_2\)

Ta có: \(n_{Fe_3O_4}=\dfrac{1}{4}n_{CO_2\left(p/ứ\right)}=0,01\left(mol\right)\) \(\Rightarrow m_{Fe_3O_4}=0,01\cdot252=2,52\left(g\right)\)

Mặt khác: \(n_{Fe}=0,07-0,01\cdot3=0,04\left(mol\right)\) \(\Rightarrow m_{Fe}=0,04\cdot56=2,24\left(g\right)\)

\(\Rightarrow m_{Al_2O_3}=6,22-2,52-2,24=1,46\left(g\right)\) 

\(a.M_A=32\cdot2=64\left(\dfrac{g}{mol}\right)\)

\(b.\)

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

\(m_A=0.25\cdot64=16\left(g\right)\)

a, Ta có: \(M_A=32.2=64\left(g/mol\right)\)

b, Có: \(n_A=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

\(\Rightarrow m_A=0,25.64=16\left(g\right)\)

Bạn tham khảo nhé!

\(n_{Na_2CO_3}=\dfrac{10.6}{106}=0.1\left(mol\right)\)

\(n_{BaCl_2}=\dfrac{250\cdot20.8\%}{208}=0.25\left(mol\right)\)

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

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

\(0.25.............0.1\)

\(LTL:\dfrac{0.25}{1}>\dfrac{0.1}{1}\Rightarrow BaCl_2dư\)

\(n_{BaCO_3}=n_{Na_2CO_3}=0.1\left(mol\right)\)

\(m_{BaCO_3}=0.1\cdot197=19.7\left(g\right)\)

\(m_{\text{dung dịch sau phản ứng}}=10.6+250-19.7=240.9\left(g\right)\)

\(C\%_{NaCl}=\dfrac{0.2\cdot58.5}{240.9}\cdot100\%=4.85\%\)

a) 

Na₂CO₃+BaCl₂------>BaCO₃+2NaCl

B) n_NaCO3 =0,1(mol)

=>m_BaCo3= 0,1*(137+12+16*3)=19,7(g)

a)       Na₂CO₃ + BaCl₂ → 2NaCl + BaCO₃↓

b)   nNa₂CO₃ = \(\dfrac{10,6}{106}\) = 0,1 mol

   mBaCl₂ = \(\dfrac{250.20,8}{100}=52\) gam

  nBaCl₂ = \(\dfrac{52}{208}=0,25\) mol

                Na₂CO₃ + BaCl₂ → 2NaCl + BaCO₃↓

Ban đầu:     0,1        0,25 

Pư:             0,1         0,1             0,2         0,1

Sau pư:       0           0,15           0,2         0,1

mBaCO₃ = 0,1.197=19,7 gam

c)   mdd= 10,6+250-19,7=240,9 gam

C%_NaCl = \(\dfrac{0,2.58,5}{240,9}.100\%=4,86\%\)

Coi n CuSO4 = 1(mol) ; n H2O = a(mol)

Ta có : 

n Cu = n CuSO4 = 1(mol)

n O = 4n CuSO4 + n H2O = 4 + a(mol)

Suy ra : 

\(\dfrac{n_{Cu}}{n_O} = \dfrac{1}{4+a} = \dfrac{1}{6}\)

=> a = 2

Vậy :

m dd X = m CuSO4 + m H2O = 1.160 + 2.18 = 196(gam)

C% CuSO4 = 160/196  .100% = 81,63%

\(n_{P_2O_5}=\dfrac{14.2}{142}=0.1\left(mol\right)\)

\(P_2O_5+3H_2O\rightarrow2H_3PO_4\)

\(0.1............................0.2\)

\(m_{H_3PO_4}=0.2\cdot98=19.6\left(g\right)\)

\(m_{dd_{H_3PO_4}}=14.2+185.8=200\left(g\right)\)

\(C\%H_3PO_4=\dfrac{19.6}{200}\cdot100\%=9.8\%\)

\(V_{dd_{H_3PO_4}}=\dfrac{200}{1.1}=181.8\left(ml\right)=0.1818\left(l\right)\)

\(C_{M_{H_3PO_4}}=\dfrac{0.2}{0.1818}=1.1\left(M\right)\)

Ta có: \(n_{P_2O_5}=\dfrac{14,2}{142}=0,1\left(mol\right)\)

PT: \(P_2O_5+3H_2O\rightarrow2H_3PO_4\)

____0,1_____________0,2 (mol)

\(\Rightarrow m_{H_3PO_4}=0,2.98=19,6\left(g\right)\)

Có: m _{dd sau pư} = m_P2O5 + m_H2O = 200 (g)

\(\Rightarrow C\%_{H_3PO_4}=\dfrac{19,6}{200}.100\%=9,8\%\)

Có: V _{dd sau pư} = \(\dfrac{200}{1,1}=\dfrac{2000}{11}\left(ml\right)=\dfrac{2}{11}\left(l\right)\)

\(\Rightarrow C_{M_{H_3PO_4}}=\dfrac{0,2}{\dfrac{2}{11}}=1,1M\)

Bạn tham khảo nhé!

\(C_{M_{HCl}} = \dfrac{n}{V} = \dfrac{0,2}{0,1} = 2M\)

100ml = 0,1l

Nồng độ mol của dung dịch là: CM = nct/Vdd = 0,2/0,1 = 2( mol/l )

PTHH: \(2KClO_3\rightarrow2KCl+3O_2\)

\(1\rightarrow1,5\left(mol\right)\)

Theo phương trình: \(n_{O_2lt}=\dfrac{1.3}{2}=1,5\left(mol\right)\)

Khối lượng \(O_2\) thu được theo lý thuyết là :

\(m_{O_2lt}=1,5.32=48\left(g\right)\)

Hiệu suất phản ứng là:

\(H=\dfrac{43,2}{44}.100\%=90\%\)

n O2(tt) = 43,2/32 = 1,35(mol)

$2KClO_3 \xrightarrow{t^o} 2KCl + 3O_2$

Theo PTHH : 

n O2(lt) = 3/2 n KClO3 = 1,5(mol)

H = n O2(tt) / n O2(lt) .100% = 1,35/1,5   .100% = 90%

PTHH: \(CaCO_3\rightarrow Cao+CO_2\)

\(1mol\rightarrow1mol\)

\(100g\rightarrow65g\)

Theo bài: 4 tấn \(\rightarrow\)m tấn

Khối lượng \(CaO\)được theo lý thuyết là:

\(m_{CaOlt}\)=\(\dfrac{4.56}{100}=2,24\)(tấn)

Hiệu suất phản ứng là:

\(H=\dfrac{1,68}{22,4.}100\%=75\%\)

nAl=10/27(mol)

ta ccó pthh: 2Al+3S->Al2S3( nhiệt dộ cao)

theo ptth=> nAl2S3(lý thuyết)=1/2.nAl=\(\dfrac{1}{2}.\dfrac{10}{27}\)=\(\dfrac{5}{27}\)(mol)

=> mAl2S3(lý thuyết)=\(\dfrac{5}{27}.150=\dfrac{250}{9}\)(g)

=>H=\(\dfrac{mAL2S3\left(thucte\right)}{mAL2S3\left(lythuyet\right)}.100\%=\dfrac{25,5}{\dfrac{250}{9}}=91,8\%\)

\(n_{Al}=\dfrac{10}{27}\left(mol\right)\)

\(2Al+3S\underrightarrow{^{t^0}}Al_2S_3\)

\(\dfrac{10}{27}...........\dfrac{5}{27}\)

\(m_{Al_2S_3\left(lt\right)}=\dfrac{5}{27}\cdot150=\dfrac{250}{9}\left(g\right)\)

\(H\%=\dfrac{m_{tt}}{m_{lt}}\cdot100\%=\dfrac{25.5}{\dfrac{250}{9}}\cdot100\%=91.8\%\)

Gọi hiệu suất là

Ta có :

n CuO = 20/80 = 0,25(mol)

n Cu = n CuO(tt) = 0,25a(mol)

=> n CuO dư = 0,25 - 0,25a(mol)

$CuO + H_2 \xrightarrow{t^o} Cu + H_2O$

Suy ra :

0,25a.64 + (0,25 - 0,25a).80 = 16,8

=>a = 0,8 = 80%