Câu 1:

a, \(2Na+2H_2O\rightarrow2NaOH+H_2\)

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

b, Ta có: \(23n_{Na}+39n_K=4,18\left(1\right)\)

Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{Na}+\dfrac{1}{2}n_K=\dfrac{1,568}{22,4}=0,07\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Na}=0,08\left(mol\right)\\n_K=0,06\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Na}=0,08.23=1,84\left(g\right)\\m_K=0,06.39=2,34\left(g\right)\end{matrix}\right.\)

c, Theo PT: \(n_{NaOH}=n_{Na}=0,08\left(mol\right)\Rightarrow m_{NaOH}=0,08.40=3,2\left(g\right)\)

Câu 2:

\(n_{Na_2O}=\dfrac{12,4}{62}=0,2\left(mol\right)\)

\(n_{BaO}=\dfrac{15,3}{153}=0,1\left(mol\right)\)

PT: \(Na_2O+H_2O\rightarrow2NaOH\)

______0,2_______________0,4 (mol)

\(BaO+H_2O\rightarrow Ba\left(OH\right)_2\)

0,1___________________0,1 (mol)

⇒ m = m_NaOH + m_Ba(OH)2 = 0,4.40 + 0,1.171 = 33,1 (g)

câu 1

\(n_{H_2\left(dktc\right)}=\dfrac{V}{22,4}=\dfrac{1,568}{22,4}=0,07\left(mol\right)\)

đặt `n_(Na)=a`, `n_K=b`

\(PTHH:2Na+2H_2O->2NaOH+H_2\)

tỉ lệ            2   :       2       :       2       ;       1

n(mol)       a----->a------------>a------------>`1/2a`

\(PTHH:2K+2H_2O->2KOH+H_2\)

Tỉ lệ          2   :     2        :      2        ;   1

n(mol)       b-->b------------->b-------->`1/2b`

ta có

\(\left\{{}\begin{matrix}23a+39b=4,18\\\dfrac{1}{2}a+\dfrac{1}{2}b=0,07\end{matrix}\right.< =>\left\{{}\begin{matrix}a=0,08\\b=0,06\end{matrix}\right.\)

\(m_{Na}=n\cdot M=0,08\cdot23=1,84\left(g\right)\\ m_K=0,06\cdot39=2,34\left(g\right)\\ m_{NaOH}=n\cdot M=0,08\cdot\left(23+16+1\right)=3,2\left(g\right)\)

a) 

2NaOH + 2H₂O --> 2NaOH + H₂

2K + 2H₂O --> 2KOH + H₂

b)

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

Gọi số mol Na, K là a, b 

=> 23a + 39b = 6,2

PTHH: 2NaOH + 2H₂O --> 2NaOH + H₂

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

2K + 2H₂O --> 2KOH + H₂

b------------------------->0,5b

=> 0,5a + 0,5b = 0,1

=> a = 0,1; b = 0,1

=> \(\left\{{}\begin{matrix}\%Na=\dfrac{0,1.23}{6,2}.100\%=37,097\%\\\%K=\dfrac{0,1.39}{6,2}.100\%=62,903\%\end{matrix}\right.\)

Gọi $n_{Na} = a(mol)$

2Na + 2H₂O → 2NaOH + H₂

a...........................a..........0,5a.....(mol)

2Al + 2NaOH + 2H₂O → 2NaAlO₂ + 3H₂

..a...........a............................................1,5a....(mol)

Suy ra : $0,5a + 1,5a = \dfrac{3,36}{22,4} = 0,15 \Rightarrow a = 0,075$

Vậy :

$m = 0,075.23 + 0,075.27 + 1,35 = 5,1(gam)$

Gọi nNa=a(mol)���=�(���)

2Na + 2H₂O → 2NaOH + H₂

a...........................a..........0,5a.....(mol)

2Al + 2NaOH + 2H₂O → 2NaAlO₂ + 3H₂

..a...........a............................................1,5a....(mol)

Suy ra : 

a, \(n_{O_2}=\dfrac{4,8}{32}=0,15\left(mol\right)\)

\(n_{H_2}=\dfrac{3}{2}=1,5\left(mol\right)\)

\(\Rightarrow V_{hh}=V_{O_2}+V_{H_2}=0,15.22,4+1,5.22,4=36,96\left(l\right)\)

b, PT:  \(O_2+2H_2\underrightarrow{t^o}2H_2O\)

Xét tỉ lệ: \(\dfrac{0,15}{1}< \dfrac{1,5}{2}\), ta được H₂ dư.

Theo PT: \(n_{H_2O}=2n_{O_2}=0,3\left(mol\right)\)

\(\Rightarrow m_{H_2O}=0,3.18=5,4\left(g\right)\)

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

PT: \(2M+2H_2O\rightarrow2MOH+H_2\)

Theo PT: \(n_M=2n_{H_2}=0,4\left(mol\right)\Rightarrow M_M=\dfrac{9,2}{0,4}=23\left(g/mol\right)\)

Vậy: M là Natri (Na).

Ta có: m _{dd sau pư} = 9,2 + 5,4 - 0,2.2 = 14,2 (g)

Theo PT: \(n_{NaOH}=2n_{H_2}=0,4\left(mol\right)\Rightarrow m_{NaOH}=0,4.40=16\left(g\right)\)

Đến đây thì m _{chất tan} lại lớn hơn cả m _{dd sau pư}. Không biết đề có nhầm lẫn gì không bạn nhỉ?

Đáp án C