a, \(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

PT: \(2Na+2H_2O\rightarrow2NaOH+H_2\)

\(Na_2O+H_2O\rightarrow2NaOH\)

Theo PT: \(n_{Na}=2n_{H_2}=0,1\left(mol\right)\Rightarrow m_{Na}=0,1.23=2,3\left(g\right)\)

\(\Rightarrow m_{Na_2O}=8,5-2,3=6,2\left(g\right)\)

b, \(n_{Na_2O}=\dfrac{6,2}{62}=0,1\left(mol\right)\)

Theo PT: \(n_{NaOH}=n_{Na}+2n_{Na_2O}=0,3\left(mol\right)\Rightarrow m_{NaOH}=0,3.40=12\left(g\right)\)

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, Giả sử: \(\left\{{}\begin{matrix}n_{Fe}=x\left(mol\right)\\n_{Zn}=y\left(mol\right)\end{matrix}\right.\)

⇒ 56x + 65y = 12,1 (1)

Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

Các quá trình: 

\(Fe^0\rightarrow Fe^{+2}+2e\)

x___________ 2x (mol)

\(Zn^0\rightarrow Zn^{+2}+2e\)

y____________ 2y (mol)

\(2H^++2e\rightarrow H_2^0\)

______0,4___0,2 (mol)

Theo ĐLBT mol e, có: 2x + 2y = 0,4 (2)

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

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,1.56}{12,1}.100\%\approx46,3\%\\\%m_{Zn}\approx53,7\%\end{matrix}\right.\)

b, BTNT Fe và Zn, có: \(\left\{{}\begin{matrix}n_{FeCl_2}=n_{Fe}=0,1\left(mol\right)\\n_{ZnCl_2}=n_{Zn}=0,1\left(mol\right)\end{matrix}\right.\)

⇒ m _muối = m_FeCl2 + m_ZnCl2 = 0,1.127 + 0,1.136 = 26,3 (g)

c, BTNT H, có: \(n_{HCl}=2n_{H_2}=0,4\left(mol\right)\)

\(\Rightarrow x=C_{M_{HCl}}=\dfrac{0,4}{0,5}=0,8\left(M\right)\)

Bạn tham khảo 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!

a, \(2Na+2H_2O\rightarrow2NaOH+H_2\)

\(Na_2O+H_2O\rightarrow2NaOH\)

Ta có: \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

Theo PT: \(n_{Na}=2n_{H_2}=0,3\left(mol\right)\Rightarrow m_{Na}=0,3.23=6,9\left(g\right)\)

\(\Rightarrow m_{Na_2O}=13,1-6,9=6,2\left(g\right)\)

b, \(n_{Na_2O}=\dfrac{6,2}{62}=0,1\left(mol\right)\)

Theo PT: \(n_{NaOH}=n_{Na}+2n_{Na_2O}=0,5\left(mol\right)\)

Ta có: m _{dd sau pư} = 13,1 + 200 - 0,15.2 = 212,8 (g)

\(\Rightarrow C\%_{NaOH}=\dfrac{0,5.40}{212,8}.100\%\approx9,4\%\)

PTHH: \(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\uparrow\)

           \(Na_2O+H_2O\rightarrow2NaOH\)

Ta có: \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\) \(\Rightarrow n_{Na}=0,6\left(mol\right)\)

\(\Rightarrow\%m_{Na}=\dfrac{0,6\cdot23}{26,2}\cdot100\%\approx52,67\left(g\right)\) \(\Rightarrow\%m_{Na_2O}=47,33\%\)

Mặt khác: \(n_{Na_2O}=\dfrac{26,2-0,6\cdot23}{62}=0,2\left(mol\right)\)

Theo PTHH: \(n_{NaOH}=n_{Na}+2n_{Na_2O}=1\left(mol\right)\) \(\Rightarrow m_{NaOH}=1\cdot40=40\left(g\right)\) 

Các bạn giúp mình với

2Na + 2H2O ---> 2NaOH + H2   (1)

Na2O + H2O ---> 2NaOH           (2)

 a) nH2 = 0,3 (mol)

Theo pthh (1) : nNa = 2nH2 = 0,6 (mol)

=> mNa = 0,6.23 = 13,8 (g)

=> mNa2O = 26,2 - 13,8 = 12,4 (g)

=> nNa2O = 0,2 (mol)

BTNa : nNaOH = nNa + 2nNa2O = 0,6 + 2.0,2 = 1 (mol)

=> mNaOH = 1.40 = 40(g)

b) %mNa = 13,8.100%/26,2 = 52,67%

%mNa2O = 100% - 52,67% = 47,33%