\(a) n_{Na} = 2a(mol) \to n_{Na_2O} = 3a(mol)\\ \Rightarrow 2a.23 + 3a.62 = 23,2\\ \Rightarrow a = 0,1\\ n_{Na} = 0,2(mol) ; n_{Na_2O} = 0,3(mol)\\ 2Na + 2H_2O \to 2NaOH + H_2\\ Na_2O + H_2O \to 2NaOH\\ n_{H_2} = \dfrac{1}{2}n_{Na} = 0,1(mol) \Rightarrow V = 0,1.22,4 = 2,24(lít)\\ b) n_{NaOH} = n_{Na} + 2n_{Na_2O} = 0,2 + 0,3.2 = 0,8(mol)\\ m_{NaOH} = 0,8.40 =32(gam) \)
PTHH: \(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\uparrow\) (1)
\(Na_2O+H_2O\rightarrow2NaOH\) (2)
a) Gọi \(n_{Na}=a\left(mol\right)\) \(\Rightarrow n_{Na_2O}=\dfrac{3}{2}a\left(mol\right)\)
\(\Rightarrow23a+62\cdot\dfrac{3}{2}a=23,2\) \(\Rightarrow a=n_{Na}=0,2\left(mol\right)\)
\(\Rightarrow n_{H_2}=0,1\left(mol\right)\) \(\Rightarrow V_{H_2}=0,1\cdot22,4=2,24\left(l\right)\)
b) Ta có: \(\left\{{}\begin{matrix}n_{Na}=0,2\left(mol\right)\\n_{Na_2O}=0,3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{NaOH\left(1\right)}=0,2\left(mol\right)\\n_{NaOH\left(2\right)}=0,6\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow n_{NaOH}=0,8\left(mol\right)\) \(\Rightarrow m_{NaOH}=0,8\cdot40=32\left(g\right)\)
a, PT: \(2Na+2H_2O\rightarrow2NaOH+H_2\)
\(Na_2O+H_2O\rightarrow2NaOH\)
Giả sử: \(\left\{{}\begin{matrix}n_{Na}=x\left(mol\right)\\n_{Na_2O}=y\left(mol\right)\end{matrix}\right.\)
⇒ 23x + 62y = 23,2 (1)
Mà: nNa : nNa2O = 2:3
⇒ 3x - 2y = 0 (2)
Từ (1) và (2) ⇒ x = 0,2 (mol); y = 0,3 (mol)
Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{Na}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)
b, Theo PT: \(n_{NaOH}=n_{Na}+2n_{Na_2O}=0,8\left(mol\right)\)
\(\Rightarrow m_{NaOH}=0,8.40=32\left(g\right)\)
Bạn tham khảo nhé!
