\(a,n_{P_2O_5}=\dfrac{8,52}{142}=0,06(mol)\\ b,n_{CaCO_3}=\dfrac{150}{100}=1,5(mol)\\ c,n_{Cu}=\dfrac{9.10^{-20}}{6.10^{-23}}=1500(mol)\\ d,n_{Na_2O}=\dfrac{15.10^{-22}}{6.10^{-23}}=25(mol)\\ e,n_{CO_2}=\dfrac{7,437}{24,79}=0,3(mol)\\ f,n_{H_2}=\dfrac{10,08}{22,4}=0,45(mol)\)

Ta có: \(n_{H_2}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\)

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

a, \(n_{Fe}=n_{H_2}=0,45\left(mol\right)\Rightarrow m_{Fe}=0,45.56=25,2\left(g\right)\)

b, \(n_{HCl}=2n_{H_2}=0,9\left(mol\right)\) \(\Rightarrow C_{M_{HCl}}=\dfrac{0,9}{0,15}=6\left(M\right)\)

c, \(FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_2+2NaCl\)

\(4Fe\left(OH\right)_2+O_2\underrightarrow{t^o}2Fe_2O_3+4H_2O\)

Theo PT: \(n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe\left(OH\right)_2}=\dfrac{1}{2}n_{FeCl_2}=\dfrac{1}{2}n_{H_2}=0,225\left(mol\right)\)

\(\Rightarrow m_{Fe_2O_3}=0,225.160=36\left(g\right)\)

nCO2=0,12(mol)

nNaOH=0,2(mol)

Ta có: 1< nNaOH/nCO2=0,2/0,12=1,667<2

=> Sp thu được hỗn hợp 2 muối Na2CO3 và NaHCO3

Đặt nCO2(1), (2) lần lượt là a, b(mol) (a,b>0)

PTHH: 2 NaOH + CO2 -to-> Na2CO3 + H2O (1)

2a_________a_________a(mol)

NaOH + CO2 -> NaHCO3 (2)

b_____b______b(mol)

Từ (1), (2) ta có hpt:

\(\left\{{}\begin{matrix}2a+b=0,2\\a+b=0,12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,08\\b=0,04\end{matrix}\right.\)

Vddsau=VddNaOH=0,2(l)

=> CMddNaHCO3= 0,04/0,2=0,2(M)

CMddNa2CO3=0,08/0,2=0,4(M)

\(n_{CO_2}=0,12mol\\ n_{NaOH}=0,2mol\\ T=\frac{n_{NaOH}}{n_{CO_2}}=\frac{0,2}{0,12}=1,66// \to Na2CO_3; NaHCO_3// 2NaOH+CO_2 \to Na_2CO_3\\ NaoH+CO_2 \to NaHCO_3\\ n_{Na_2CO_3}=a(mol)\\ n_{NaHCO_3}=b(mol)\\ n_{NaOH}=2a+b=0,2(1)\\ n_{CO_2}=a+b=0,12(2)\\ (1)(2)\\ a=0,08; b=0,04\\ CM_{Na_2CO_3}=\frac{0,08}{0,2}=0,4(mol)\\ CM_{NaHCO_3}=\frac{0,04}{0,2}=0,2(mol)\\ \)

Gọi số mol CO₂ và SO₂ là a, b (mol)

= >\(\left\{{}\begin{matrix}n_{khí}=a+b=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\\dfrac{44a+64b}{a+b}=29,5.2=59\end{matrix}\right.\) => \(\left\{{}\begin{matrix}a=0,1\left(mol\right)\\b=0,3\left(mol\right)\end{matrix}\right.\)

\(n_{NaOH}=1.0,4=0,4\left(mol\right)\)

PTHH: NaOH + CO₂ --> NaHCO₃

________0,1<----0,1------->0,1_______(mol)

NaOH + SO₂ --> NaHSO3

_0,3<----0,3-------->0,3_____________(mol)

=> \(\left\{{}\begin{matrix}C_{M\left(NaHCO_3\right)}=\dfrac{0,1}{0,4}=0,25M\\C_{M\left(NaHSO_3\right)}=\dfrac{0,3}{0,4}=0,75M\end{matrix}\right.\)

Bạn xem lại đề !

\(n_{H_2S}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ n_{NaOH}=0,25.2=0,5\left(mol\right)\)

Xét \(T=\dfrac{0,5}{0,4}=1,25\) => Tạo cả 2 muối Na₂S và NaHS

PTHH: 

2NaOH + H₂S ---> Na₂S + 2H₂O

0,5------->0,25------>0,25

Na₂S + H₂S ---> 2NaHS 

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

=> m_muối = 0,3.56 + (0,25 - 0,15).78 = 24,6 (g)

nSO₂=2,24/22.4=0,1 mol

a)

nNaOH=0,5*1=0,5mol

Đặt T=nNaOH/nSO2=0,5/0,1=5

Vì T> 2--->PT tạo muối trung hoà Na₂SO₃( NaOH dư)

 PTHH: SO₂ + 2NaOH----> Na2SO3 + H2O

    ----->nNa2SO3=nSO2=0,1 mol

------>mNa2SO3=0.1*126=12,6g

b)

nNaOH=0.15*1=0.15 mol

Đặt T=nNaOH/ nSO2=0,15/0,1=1,5

Vì 1<T<2----> P/Ứ tạo 2 muối là Na2cO3 và NaHO3

PTHH

          SO2 + 2NaOH------> Na2CO3 + H2O

            a--------- 2a    ---------  a

           SO2 + NaOH-----> NaHCO3

                b           b                       b 

Gọi số mol của N2SO3 và NaHSO3 lần lượt là  a và b

   Ta có: \(\begin{cases}a+b=0,1\\2a+b=0,15\end{cases}\)

-\(\Rightarrow\) a=0,05 mol ; b=0,05 mol

----->m Na2SO3= 0,05*126=6.3g

        mNaHSO3=0,05*72=3,6g

 không biết đúng không nữa!!!

a)

$2NaOH  + SO_2 \to Na_2SO_3 + H_2O$

b)

n SO2 = 1,12/22,4 = 0,05(mol)

n NaOH = 0,2.1,5 = 0,3(mol)

Ta thấy : 

n NaOH / n SO2 = 0,3 /0,05= 6 > 2 nên NaOH dư

n NaOH pư = 2n SO2 = 0,1(mol)

=> n NaOH dư = 0,3 - 0,1 = 0,2(mol)

n Na2SO3 = n SO2 = 0,05(mol)

Vậy : 

CM NaOH dư = 0,2/0,2 = 1M

CM Na2SO3 = 0,05/0,2 = 0,25M

\(a,\left\{{}\begin{matrix}n_C=\dfrac{4}{12}=0,25\left(mol\right)\\n_P=\dfrac{62}{31}=2\left(mol\right)\\n_{Fe}=\dfrac{42}{56}=0,75\left(mol\right)\end{matrix}\right.\\ b,\left\{{}\begin{matrix}n_{H_2O}=\dfrac{3,6}{18}=0,2\left(mọl\right)\\n_{CO_2}=\dfrac{95,48}{44}=2,17\left(mol\right)\\n_{NaCl}=\dfrac{14,625}{58,5}=0,25\left(mol\right)\end{matrix}\right.\)

\(c,\left\{{}\begin{matrix}n_{O_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\n_{CO}=\dfrac{12}{24}=0,5\left(mol\right)\end{matrix}\right.\)

a. \(n_C=\dfrac{4}{12}=\dfrac{1}{3}\left(mol\right)\)

\(n_P=\dfrac{62}{31}=2\left(mol\right)\)

\(n_{Fe}=\dfrac{42}{56}=0,75\left(mol\right)\)

b. \(n_{H_2O}=\dfrac{3.6}{18}=0,2\left(mol\right)\)

\(n_{CO_2}=\dfrac{95.48}{44}=2,17\left(mol\right)\)

\(n_{NaCl}=\dfrac{14.625}{58,5}=0,25\left(mol\right)\)

c. \(n_{O_2}=\dfrac{8.96}{22,4}=0,4\left(mol\right)\)

\(n_{H_2}=\dfrac{5.6}{22,4}=0,25\left(mol\right)\)

\(n_{CO}=\dfrac{12}{24}=0,5\left(mol\right)\)