Đáp án:
1. \(m_{H_2O\ \text{dư}}=18\ (gam)\\ m_{NaOH}=40 \ (gam)\)
2. K
3. V = 4,48 l
4. Mg
Giải thích các bước giải:
1.
\(n_{Na_2O}=\dfrac{31}{62}=0,5 \ (mol)\\ n_{H_2O}=\dfrac{27}{18}=1,5\ (mol)\\ Na_2O+H_2O\to 2NaOH\\ \dfrac{0,5}{1}<\dfrac{1,5}{1}\)
Suy ra sau phản ứng \(Na_2O\) hết, \(H_2O\)
\(n_{NaOH}=2n_{Na_2O}=0,5\times 2=1\ (mol)\\ \Rightarrow m_{NaOH}=40\times 1=40 \ (gam)\)
\(n_{H_2O\ \text{phản ứng}}=n_{Na_2O}=0,5\ \text{(mol)}\\ \Rightarrow n_{H_2O\ \text{dư}}=1,5-0,5=1 \ (mol)\\ \Rightarrow m_{H_2O\ \text{dư}}=1\times 18=18 \ (gam)\)
2.
\(n_{H_2O}=\dfrac{3,6}{18}=0,2\ \text{mol}\\ A+H_2O\to AOH+\dfrac 12H_2\\ n_A=n_{H_2O}=0,2\ (mol)\\ \Rightarrow M_A=\dfrac{7,8}{0,2}=39\\ \Rightarrow A:\ K\)
3.
\(n_{Zn}=\dfrac{13}{65}=0,2\ (mol)\\ Zn+2HCl\to ZnCl_2+H_2\\ \dfrac{0,2}{1}<\dfrac{0,5}{2}\)
\(\Rightarrow\) Sau phản ứng Zn hết, HCl dư
\(\Rightarrow n_{H_2}=n_{Zn}=0,2 \ (mol)\\ \Rightarrow V_{H_2}=0,2\times 22,4=4,48 \ (l)\)
4.
\(A+H_2SO_4\to ASO_4+H_2\\ n_{H_2}=\dfrac{4,48}{22,4}=0,2 \ (mol)\\ n_A=n_{H_2}=0,2\ (mol)\\ \Rightarrow M_A=\dfrac{4,8}{0,2}=24\\ \Rightarrow A: \ Mg\)
