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\(n_{Na}=\dfrac{4.6}{23}=0.2\left(mol\right)\)
\(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\)
\(0.2......................0.2.......0.1\)
\(V_{H_2}=0.1\cdot22.4=2.24\left(l\right)\)
Dung dịch X : NaOH
\(m_{dd_X}=4.6+200-0.1\cdot2=204.4\left(g\right)\)
\(C\%_{NaOH}=\dfrac{0.2\cdot40}{204.4}\cdot100\%=3.9\%\%\)
nZn = 13/65 = 0.2 (mol)
Zn + H2SO4 => ZnSO4 + H2
0.2......0.2..........................0.2
VH2 = 0.2*22.4 = 4.48 (l)
C%H2SO4 = 0.2*98/200 * 100% = 9.8 %
nCuO = 8/80 = 0.1 (mol)
CuO + H2 -to-> Cu + H2O
0.1......0.1...........0.1
=> H2 dư
mCu = 0.1*64 = 6.4 (g)
â) nZn=0,2(mol)
PTHH: Zn + H2SO4 -> ZnSO4 + H2
0,2_____0,2______0,2_____0,2(mol)
=> V(H2,đktc)=0,2.22,4=4,48(l)
b) C%ddH2SO4= [(98.0,2)/200)].100=9,8%
c) nCuO=0,1(mol)
PTHH: CuO + H2 -to-> Cu + H2O
Ta có: 0,1/1 < 0,2/1
=> H2 dư, CuO hết, tính theo nCuO
=> nCu=nCuO=0,1(mol)
=>mCu=6,4(g)
n Zn= 19,5/65=0,3 (mol).
PTPƯ: Zn(0.3) + HCl(0.6) ----> ZnCl2(0.3) + H2(0,3)
mHCl=0,6.36.5=21.9(g)
a) C%HCl= 21.9/300.100%=7,3%
b) VH2=0,3.22,4=6,72(lít)
c) mH2=0,3.2=0,6(g)
mZnCl2=0,3.136=40,8(g)
mddZnCl2 =(19,5+300)-0,6=318,9(g)
C%=mZnCl2/mddZnCl2.100= 40,8/318,9.100=12,793%
\(n_{Na}=\dfrac{2,3}{23}=0,1\left(mol\right)\)
PTHH: 2Na + 2H2O --> 2NaOH + H2
0,1-------------->0,1---->0,05
=> VH2 = 0,05.22,4 = 1,12 (l)
mdd sau pư = 2,3 + 197,8 - 0,05.2 = 200 (g)
=> \(C\%=\dfrac{0,1.40}{200}.100\%=2\%\)
\(V_{dd}=\dfrac{200}{1,08}=\dfrac{5000}{27}\left(ml\right)=\dfrac{5}{27}\left(l\right)\)
=> \(C_M=\dfrac{0,1}{\dfrac{5}{27}}=0,54M\)
Ta có : \(n_K=\dfrac{m}{M}=\dfrac{3,9}{39}=0,1\) (mol)
\(n_{H_2O}=\dfrac{m}{M}=\dfrac{96,2}{18}=5,34\)(mol)
Phương trình hóa học :
2K + 2H2O ---> 2KOH + H2
2 : 2 : 2 : 1
Nhận thấy \(\dfrac{n_K}{n_{H_2O}}=\dfrac{0,1}{5,34}< \dfrac{2}{2}\)
=> Kali hết , nước dư
=> \(n_{H_2}=\dfrac{n_K}{2}=0,05\) (mol)
=> Thể tích khí H2 : V = n.22,4 = 0,05.22,4 = 1,12(l)
Lại có \(n_{KOH}=0,1\) (mol) => \(m_{KOH}=0,1.56=5,6\) (g)
\(m_{H_2}=0,05.2=0,1\left(g\right)\)
Nồng độ phần trăm của Base thu được :
\(C\%=\dfrac{m_{KOH}}{m_{dd}-m_{H_2}}=\dfrac{5,6}{96,2+3,9-0,1}=0,056=5,6\%\)
nNa = 9.2/23 = 0.4 (mol)
2Na + 2H2O => 2NaOH + H2
0.4.........................0.4.......0.2
VH2 = 0.2 * 22.4 = 4.48 (l)
mNaOH = 0.4 * 40 = 16 (g)
mdd = 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: \(m_{dd\left(saup/ứ\right)}=m_{Na}+m_{H_2O}-m_{H_2}=108,8\left(g\right)\)
\(\Rightarrow C\%_{NaOH}=\dfrac{16}{108,8}\cdot100\%\approx14,71\%\)
\(n_{Mg}=\frac{m}{M}=\frac{9,6}{24}=0,4mol\)
PTHH: \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\)
1 : 1 : 1 : 1 mol
0,4 0,4 0,4 0,4 mol
a. \(m_{MgSO_4}=n.M=0,4.\left(24+32+16.4\right)=48g\)
b. \(V_{H_2}=n.22,4=0,4.22,4=8,96l\)
c. \(n_{Fe_2O_3}=\frac{m}{M}=\frac{64}{56.2}+16.3=0,4mol\)
PTHH: \(3H_2+Fe_{2O_3}\rightarrow2Fe+3H_2O\left(ĐK:t^o\right)\)
3 : 1 : 2 : 3 mol
1, 7 0,4 0,8 1,2 mol
\(m_{Fe}=n.M=0,8.56=44,8g\)
Ta có: \(n_{Na}=\dfrac{9,2}{23}=0,4\left(mol\right)\)
\(PTHH:2Na+2H_2O--->2NaOH+H_2\uparrow\)
a. Theo PT: \(n_{H_2}=\dfrac{1}{2}.n_{Na}=\dfrac{1}{2}.0,4=0,2\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,2.22,4=4,48\left(lít\right)\)
b. Theo PT: \(n_{NaOH}=n_{Na}=0,4\left(mol\right)\)
\(\Rightarrow m_{NaOH}=0,4.40=16\left(g\right)\)
Ta có: \(m_{dd_{NaOH}}=9,2+200-0,2.2=208,8\left(g\right)\)
\(\Rightarrow C_{\%_{NaOH}}=\dfrac{16}{208,8}.100\%=7,66\%\)