a) PTHH: Fe + 2 HCl -> FeCl2 + H2

nH2= 0,1(mol)

-> nFe= nFeCl2=nH2=0,1(mol)

=>mFeCl2=127.0,1=12,7(g)

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

mFeCl3= m(hỗn hợp muối)- mFeCl2= 45,2- 12,7= 32,5(g)

b) => nFeCl3= 0,2(mol)

=> nFe2O3= nFeCl3/2= 0,2/2= 0,1(mol)

=> m(hỗn hợp ban đầu)= mFe+ mFe2O3= 0,1. 56+ 0,1.160=21,6(g)

a) PTHH: Fe + 2 HCl -> FeCl2 + H2

nH2= 0,1(mol)

-> nFe= nFeCl2=nH2=0,1(mol)

=>mFeCl2=127.0,1=12,7(g)

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

mFeCl3= m(hỗn hợp muối)- mFeCl2= 45,2- 12,7= 32,5(g)

b) => nFeCl3= 0,2(mol)

=> nFe2O3= nFeCl3/2= 0,2/2= 0,1(mol)

=> m(hỗn hợp ban đầu)= mFe+ mFe2O3= 0,1. 56+ 0,1.160=21,6(g)

Số mol của 2,24 lít khí H2:

\(n_{H_2}=\dfrac{V}{22,4}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

PTHH:

\(2Na+2H_2O\rightarrow2NaOH+H_2\uparrow\left(1\right)\)

2     :    2        :    2             : 1 

0,1->   0,1     :   0,1           :  0,05(mol)

\(Na_2O+H_2O\rightarrow2NaOH\left(2\right)\)

Khối lượng của 0,2 mol Na:

\(m_{Na}=n.M=0,2.23=4,6\left(g\right)\)

Do hỗn hợp A gồm Na và H2O nên ta có:

\(m_{Hỗnhợp}=m_{Na}+m_{Na_2O}\\ \Rightarrow m_{Na_2O}=m_{Hỗnhợp}-m_{Na}\\ \Rightarrow m_{Na_2O}=12,4-4,6\\ \Rightarrow m_{Na_2O}=7,8\left(g\right)\)

a) Gọi số mol Na, Ca là a, b (mol)

=> 23a + 40b = 8,3 (1)

PTHH: 2Na + 2H₂O --> 2NaOH + H₂

              a--------------->a------>0,5a

             Ca + 2H₂O --> Ca(OH)₂ + H₂

              b--------------->b--------->b

=> \(n_{H_2}=0,5a+b=\dfrac{4,48}{22,4}=0,2\left(mol\right)\) (2)

(1)(2) => a = 0,1; b = 0,15

=> \(\left\{{}\begin{matrix}\%m_{Na}=\dfrac{0,1.23}{8,3}.100\%=27,71\%\\\%m_{Ca}=\dfrac{0,15.40}{8,3}.100\%=72,29\%\end{matrix}\right.\)

b) \(\left\{{}\begin{matrix}m_{NaOH}=0,1.40=4\left(g\right)\\m_{Ca\left(OH\right)_2}=0,15.74=11,1\left(g\right)\end{matrix}\right.\)

\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ 2Na+2H_2O\rightarrow2NaOH+H_2\\ Ca+2H_2O\rightarrow Ca\left(OH\right)_2+H_2\\ Đặt:n_{Na}=a\left(mol\right);n_{Ca}=b\left(mol\right)\left(a,b>0\right)\\ \Rightarrow\left\{{}\begin{matrix}23a+40b=8,3\\0,5a+b=0,2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,15\end{matrix}\right.\\ b,\Rightarrow\%m_{Ca}=\dfrac{0,15.40}{8,3}.100\approx72,289\%\\ \Rightarrow\%m_{Na}\approx27,711\%\\ b,n_{NaOH}=a=0,1\left(mol\right)\\ n_{Ca\left(OH\right)_2}=b=0,15\left(mol\right)\\ m_{bazo}=m_{NaOH}+m_{Ca\left(OH\right)_2}=40.0,1+74.0,15=15,1\left(g\right)\)

2Na+2H₂O->2NaOH+H₂

x------------------------------0,5x

Ca+2H₂O->Ca(OH)₂+H₂

y-------------------------------y 

Ta có :

\(\left\{{}\begin{matrix}23x+40y=8,3\\0,5x+y=0,2\end{matrix}\right.=>\left\{{}\begin{matrix}x=0,1\\y=0,15\end{matrix}\right.\)

=>%mNa=\(\dfrac{0,1.23}{8,3}.100=27,71\%\)

=>%mCa=72,29%

b)m bazo=0,1.40+0,15.74=15,1g

\(2Na+2H_2O\rightarrow2NaOH+H_2\)

\(2K+2H_2O\rightarrow2KOH+H_2\)

Gọi: \(\left\{{}\begin{matrix}n_{Na}=x\left(mol\right)\\n_K=y\left(mol\right)\end{matrix}\right.\) ⇒ 23x + 39y = 12,4 (1)

Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{Na}+\dfrac{1}{2}n_K=\dfrac{1}{2}x+\dfrac{1}{2}y=\dfrac{2,24}{22,4}=0,1\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=-0,2875\\y=0,4875\end{matrix}\right.\)

Đến đây thì ra số mol âm, bạn xem lại đề nhé.

\(2Na+2H_2O\rightarrow2NaOH+H_2\\ Na_2O+H_2O\rightarrow2NaOH\\ n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ n_{Na}=2.0,3=0,6\left(mol\right)\\ a,m_{Na}=0,6.23=13,8\left(g\right)\\ m_{Na_2O}=26,2-13,8=12,4\left(g\right)\\b, n_{Na_2O}=\dfrac{12,4}{62}=0,2\left(mol\right)\\ n_{NaOH\left(tổng\right)}=n_{Na}+2.n_{Na_2O}=0,6+\dfrac{12,4}{62}=0,8\left(mol\right)\\ m_{c.tan}=m_{NaOH}=0,8.40=32\left(g\right)\\ c,m_{ddNaOH}=m_{hh}+m_{H_2O}-m_{H_2}=26,2+200-0,3.2=225,6\left(g\right)\\ C\%_{ddNaOH}=\dfrac{32}{225,6}.100\approx14,185\%\)

Anh có làm rồi nha!

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

PT: \(2K+2H_2O\rightarrow2KOH+H_2\)

\(K_2O+H_2O\rightarrow2KOH\)

Theo PT: \(n_K=2n_{H_2}=0,2\left(mol\right)\Rightarrow m_K=0,2.39=7,8\left(g\right)\)

\(\Rightarrow m_{K_2O}=12,4-7,8=4,6\left(g\right)\)

a) Fe + 2HCl --> FeCl₂ + H₂

FeO + 2HCl --> FeCl₂ + H₂O

b) \(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

______0,5<-1<------0,5<---0,5

=> m_Fe = 0,5.56 = 28 (g)

=> \(\left\{{}\begin{matrix}\%Fe=\dfrac{28}{100}.100\%=28\%\\\%FeO=100\%-28\%=72\%\end{matrix}\right.\)

c) \(n_{FeO}=\dfrac{72}{72}=1\left(mol\right)\)

PTHH: FeO + 2HCl --> FeCl₂ + H₂O

______1---->2

=> m_HCl = (1+2).36,5 = 109,5 (g)

=> \(m_{ddHCl}=\dfrac{109,5.100}{30}=365\left(g\right)\)

=> \(V_{ddHCl}=\dfrac{365}{1,15}=317,39\left(ml\right)\)