Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)
$Na_2O + H_2O \to 2NaOH + H_2$
$2Na + 2H_2O \to 2NaOH + H_2$
b)
$n_{Na} = 2n_{H_2} = 0,05.2 = 0,1(mol)$
$\Rightarrow n_{Na_2O} = \dfrac{14,7 - 0,1.23}{62} = 0,2(mol)$
$n_{NaOH} = n_{Na} + 2n_{Na_2O} = 0,5(mol)$
$C_{M_{NaOH}} = \dfrac{0,5}{0,2} = 2,5M$
c)
$2NaOH + H_2SO_4 \to Na_2SO_4 + 2H_2O$
$n_{H_2SO_4} = \dfrac{1}{2}n_{NaOH} = 0,25(mol)$
$m_{dd\ H_2SO_4} = \dfrac{0,25.98}{20\%} = 122,5(gam)$
$V_{dd\ H_2SO_4} = \dfrac{122,5}{1,4} = 87,5(ml)$
a)
$n_{Na_2O} = \dfrac{15,5}{62} = 0,25(mol)$
$Na_2O + H_2O \to 2NaOH$
$n_{NaOH} = 2n_{Na_2O} = 0,5(mol)$
$C_{M_{NaOH}} = \dfrac{0,5}{0,5} = 1M$
b)
$2NaOH + H_2SO_4 \to Na_2SO_4 + 2H_2O$
$n_{H_2SO_4} = \dfrac{1}{2}n_{NaOH} = 0,25(mol)$
$\Rightarrow m_{dd\ H_2SO_4} = \dfrac{0,25.98}{20\%} = 122,5(gam)$
$\Rightarrow V_{dd\ H_2SO_4} = \dfrac{122,5}{1,14} = 107,46(ml)$
\(n_{H_2}=\dfrac{1.12}{22.4}=0.05\left(mol\right)\)
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
\(0.05.......0.05......0.05...........0.05\)
\(m_{Zn}=0.05\cdot65=3.25\left(g\right)\)
\(C_{M_{H_2SO_4}}=\dfrac{0.05}{0.2}=0.25\left(M\right)\)
\(m_{ZnSO_4}=0.05\cdot161=8.05\left(g\right)\)
a, \(Na_2O+H_2O\rightarrow2NaOH\)
\(n_{Na_2O}=\dfrac{15,5}{62}=0,25\left(mol\right)\)
\(n_{NaOH}=2n_{Na_2O}=0,5\left(mol\right)\)
\(\Rightarrow C_{M_{NaOH}}=\dfrac{0,5}{0,5}=1\left(M\right)\)
b, \(H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\)
\(n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,25\left(mol\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,25.98}{20\%}=122,5\left(g\right)\)
\(\Rightarrow V_{ddH_2SO_4}=\dfrac{122,5}{1,14}\approx107,46\left(ml\right)\)
\(n_{H2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
Pt : \(Fe+H_2SO_4\rightarrow FeSO_4+H_2|\)
1 1 1 1
0,25 0,25 0,25
a) \(n_{Fe}=\dfrac{0,25.1}{1}=0,25\left(mol\right)\)
⇒ \(m_{Fe}=0,25.56=14\left(g\right)\)
b) \(n_{H2SO4}=\dfrac{0,25.1}{1}=0,25\left(mol\right)\)
200ml = 0,2l
\(C_{M_{ddH2SO4}}=\dfrac{0,25}{0,2}=1,25\left(M\right)\)
Chúc bạn học tốt
\(n_{Zn}=\dfrac{4,55}{65}=0,07(mol)\\ Zn+2HCl\to ZnCl_2+H_2\\ a,n_{HCl}=0,14(mol)\\ \Rightarrow C_{M_{HCl}}=\dfrac{0,14}{0,2}=0,7M\\ b,n_{H_2}=0,07(mol)\\ \Rightarrow V_{H_2}=0,07.22,4=1,568(l)\\ c,n_{ZnCl_2}=0,07(mol)\\ \Rightarrow m_{ZnCl_2}=0,07.136=9,52(g)\\ c,ZnCl_2+2AgNO_3\to 2AgCl\downarrow+Zn(NO_3)_2\)
\(m_{dd_{ZnCl_2}}=200.0,8+4,55-0,07.2=164,41(g)\\ n_{AgCl}=0,14(mol);n_{Zn(NO_3)_2}=0,07(mol)\\ \Rightarrow C\%_{Zn(NO_3)_2}=\dfrac{0,07.189}{164,41+200-0,14.143,5}.100\%=3,84%\)