a: 2Na+2H2O->2NaOH+H2

0,1                         0,1     0,1

nNa=2,3/23=0,1mol

=>nNaOH=0,1mol

m=0,1*40=4(g)

b: V=0,1*22,4=2,24(lít)

\(a.2Na+2H_2O\rightarrow2NaOH+H_2\\ b.n_{Na}=\dfrac{9,2}{23}=0,4\left(mol\right)\\ n_{H_2}=\dfrac{1}{2}n_{Na}=0,2\left(mol\right)\\ \Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\\ n_{NaOH}=n_{Na}=0,4\left(mol\right)\\ \Rightarrow m_{NaOH}=0,4.40=16\left(g\right)\\ c.H_2+CuO-^{t^o}\rightarrow Cu+H_2O\\ n_{CuO}=\dfrac{12}{80}=0,15\left(mol\right)\\ LTL:\dfrac{0,2}{1}>\dfrac{0,15}{1}\Rightarrow H_2dưsauphảnứng\\ n_{H_2\left(pứ\right)}=n_{CuO}=0,15\left(mol\right)\\ \Rightarrow n_{H_2\left(dư\right)}=0,2-0,15=0,05\left(mol\right)\\ \Rightarrow m_{H_2\left(Dư\right)}=0,05.2=0,1\left(g\right)\)

a) Zn + 2HCl →ZnCl₂  + H₂

b) nZn = 6,5/65 = 0,1 mol . Theo tỉ lệ pư => nH₂ = nZn = nZnCl₂ =0,1 mol <=> V_H2(đktc) = 0,1.22,4 = 2,24 lít.

c) mZnCl₂ = 0,1 . 136 = 13,6 gam

d) nHCl =2nZn = 0,2 mol => mHCl = 0,2.36,5= 7,3 gam

Cách 2: áp dụng định luật BTKL => mHCl = mZnCl₂ + mH₂ - mZn 

<=> mHCl = 13,6 + 0,1.2 - 6,5 = 7,3 gam

n_Na = 9.2/23 = 0.4 (mol) 

2Na + 2H₂O => 2NaOH + H₂

0.4.........................0.4.......0.2

V_H2 = 0.2 * 22.4 = 4.48 (l) 

m_NaOH = 0.4 * 40 = 16 (g) 

m_dd = 9.2 + 100 - 0.2 * 2 = 108.8 (g) 

C% _NaOH = 16 / 108.8 * 100% = 14.71%

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

Ta có: \(n_{Na}=\dfrac{9,2}{23}=0,4\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{NaOH}=0,4\left(mol\right)\\n_{H_2}=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{NaOH}=0,4\cdot40=16\left(g\right)\\V_{H_2}=0,2\cdot22,4=4,48\left(l\right)\\m_{H_2}=0,2\cdot2=0,4\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd\left(saup/ứ\right)}=m_{Na}+m_{H_2O}-m_{H_2}=108,8\left(g\right)\)

\(\Rightarrow C\%_{NaOH}=\dfrac{16}{108,8}\cdot100\%\approx14,71\%\)

nNa=9,2/23=0,4(mol).

PTPƯ: 2Na (0,4)   +      2 H2O(0,4)----> 2NaOH (0,4)+    H2(0,2).

a)VH2= 0,2.22,4=4,48(lít).

mNaOH= 0,4.40=16(g).

b)mH2=0,2.2=0,4(mol).

mdd(sau pư)=mNa+mH2O-mH2=9,2+100-0,4=108,8(g)

C%ddNaOH=16/108,8.100=14,705%.

Theo gt ta có: $n_{Zn}=0,1(mol)$

a, $Zn+2HCl\rightarrow ZnCl_2+H_2$

b, Ta có: $n_{H_2}=0,1(mol)\Rightarrow V_{H_2}=2.24(l)$

c, Ta có: $n_{HCl}=2.n_{Zn}=0,2(mol)\Rightarrow m_{HCl}=7,3(g)$

\(a)\\ Zn + 2HCl \to ZnCl_2 + H_2\\ b)\\ n_{H_2} = n_{Zn} = \dfrac{6,5}{65} = 0,1(mol)\\ \Rightarrow V_{H_2} = 0,1.22,4 = 2,24(lít)\\ c)\\ n_{HCl} = 2n_{Zn} = 0,1.2 = 0,2(mol)\\ \Rightarrow m_{HCl} = 0,2.36,5 = 7,3\ gam\)

a, PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

b, Ta có: \(n_{Fe}=\dfrac{14}{56}=0,25\left(mol\right)\)

Theo PT: \(n_{HCl}=2n_{Fe}=0,5\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,25.36,5=18,25\left(g\right)\)

c, Theo PT: \(n_{H_2}=n_{Fe}=0,25\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,25.22,4=5,6\left(l\right)\)

\(a.Fe+2HCl\rightarrow FeCl_2+H_2\\b.n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\\ n_{H_2}=n_{Fe}=0,1\left(mol\right)\\ \Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\\ c.n_{FeCl_2}=n_{Fe}=0,1\left(mol\right)\\ m_{FeCl_2}=0,1.127=12,7\left(g\right) \)

a.b.c.\(n_{Al}=\dfrac{2,7}{27}=0,1mol\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

0,1                         0,1      0,15   ( mol )

\(V_{H_2}=0,15.22,4=3,36l\)

\(m_{AlCl_3}=0,1.133,5=13,35g\)

d.\(Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\)

                  0,15              0,1              ( mol )

\(m_{Fe}=0,1.56=5,6g\)