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\(a.n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ a............3a.......a.........1,5a\left(mol\right)\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ b.........2b........b.........b\left(mol\right)\\ \rightarrow\left\{{}\begin{matrix}27a+56b=5,5\\1,5a+b=0,2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,05\end{matrix}\right.\\ \rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{27.0,1}{5,5}.100\approx49,091\%\\\%m_{Fe}=\dfrac{0,05.56}{5,5}.100\approx50,909\%\end{matrix}\right.\\ b.C_{MddHCl}=\dfrac{3a+2b}{0,5}=\dfrac{3.0,1+2.0,05}{0,5}=0,8\left(M\right)\)
Bài 1:
PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
Ta có: \(n_{Mg}=\dfrac{1}{2}n_{HCl}=\dfrac{1}{2}\cdot0,11\cdot1,5=0,0825\left(mol\right)\)
\(\Rightarrow m_{Mg}=0,0825\cdot24=1,98\left(g\right)\)
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
PTHH :
\(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
0,2 0,4 0,2 0,2
\(a,V_{H_2}=0,2.22,4=4,48\left(l\right)\)
\(b,m_{HCl}=0,4.36,5=14,6\left(g\right)\)
\(m_{ddHCl}=\dfrac{14,6.100}{10}=146\left(g\right)\)
\(c,m_{MgCl_2}=0,2.95=19\left(g\right)\)
\(m_{ddMgCl_2}=4,8+146-\left(0,2.2\right)=150,4\left(g\right)\)
\(C\%_{MgCl_2}=\dfrac{19}{150,4}.100\%\approx12,63\%\)
2.
\(n_K=\dfrac{7,8}{39}=0,2\left(mol\right)\)
\(2K+2H_2O\rightarrow2KOH+H_2\uparrow\)
0,2 0,2 0,1
\(m_{KOH}=0,2.56=11,2\left(g\right)\)
\(m_{ddKOH}=7,8+100-\left(0,1.2\right)=107,6\left(g\right)\)
\(C\%=\dfrac{11,2}{107,6}.100\%\approx10,4\%\)
Ta có: \(\left\{{}\begin{matrix}n_{HCl}=0,796.0,5=0,398\left(mol\right)\\n_{H_2SO_4}=0,796.0,75=0,597\left(mol\right)\end{matrix}\right.\)
\(n_{H_2}=\dfrac{4,368}{22,4}=0,195\left(mol\right)\)
BTNT H, có: \(n_{HCl}+2n_{H_2SO_4}=2n_{H_2}+2n_{H_2O}\Rightarrow n_{H_2O}=0,601\left(mol\right)\)
Theo ĐLBT KL, có: m hh + m axit = m muối + mH2 + mH2O
⇒ m = m muối = 26,43 + 0,398.36,5 + 0,597.98 - 0,195.2 - 0,601.18 = 88,255 (g)
mH2SO4= \(\dfrac{300.7,35}{100}=22,05g\)
nH2SO4= \(\dfrac{22,05}{98}=0,225 mol\)
mHCl= \(\dfrac{200.7,3}{100}=14,6g\)
nHCl= \(\dfrac{14,6}{36,5}=0,4mol\)
H2SO4 + 2HCl → 2H2O + Cl2 ↑+ SO2 ↑
n trước pư 0,225 0,4
n pư 0,2 ← 0,4 → 0,4 → 0,2 → 0,2 mol
n sau pư dư 0,025 hết
a) mCl2= 0,2. 71= 14,2g
mSO2= 64. 0,2= 12,8g
mH2O= 18. 0,4=7,2g
mdd sau pư= 300 +200 -14,2 -12,8= 473g
C%dd H2O= \(\dfrac{7,2.100}{473}=1,52\)%
b) Mg + 2H2O → Mg(OH)2 + H2 ↑
x → 2x → x → x
Fe + 2H2O → Fe(OH)2 + H2↑
y → 2y → y → y
Gọi x,y lần lượt là số mol của Mg,Fe.
Ta có hệ phương trình:
24x + 56y = 8,7 x= \(\dfrac{5}{64}\)
⇒
2x + 2y = 0,4 y= \(\dfrac{39}{320}\)
VH2= 22,4. \((\dfrac{5}{64}+\dfrac{39}{320})\)= 4,48l
mhh MG(OH)2, Fe(OH)2= 8,7 +250 - 2.(\(\dfrac{5}{64}+\dfrac{39}{320}\)) = 2258,3g
mMg=24. \(\dfrac{5}{64}\)=1.875g
mFe= 8,7-1,875= 6,825g
a) $Fe + H_2SO_4 \to FeSO_4 + H_2$
b)
Theo PTHH : $n_{H_2SO_4} = n_{H_2} = \dfrac{224}{1000.22,4} = 0,01(mol)$
$C\%_{H_2SO_4} = \dfrac{0,01.98}{100}.100\% = 0,98\%$
$\Rightarrow x = 0,98$
c) $n_{Fe} = n_{H_2} = 0,01(mol)$
$m_{dd\ sau\ pư} = m_{Fe} + m_{dd\ H_2SO_4} - m_{H_2} = 0,01.56 + 100 - 0,01.2 = 100,54(gam)$
$C\%_{FeSO_4} = \dfrac{0,01.152}{100,54}.100\% = 1,51\%$
\(n_{Mg}=\dfrac{4.8}{24}=0.2\left(mol\right)\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(0.2.....................0.2..........0.2\)
\(V_{H_2}=0.2\cdot22.4=4.48\left(l\right)\)
\(m_{\text{dung dịch sau phản ứng }}=4.8+250-0.2\cdot2=254.4\left(g\right)\)
\(m_{MgCl_2}=0.2\cdot95=19\left(g\right)\)
\(C\%_{MgCl_2}=\dfrac{19}{254.4}\cdot100\%=7.47\%\)
Ta có: \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
a, PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
____0,2____0,4______0,2____0,2 (mol)
b, \(V_{H_2}=0,2.22,4=4,48\left(l\right)\)
c, Ta có: m dd sau pư = mMg + m dd HCl - mH2 = 4,8 + 250 - 0,2.2 = 254,4 (g)
\(\Rightarrow C\%_{MgCl_2}=\dfrac{0,2.95}{254,4}.100\%\approx7,47\%\)
Bạn tham khảo nhé!
\(n_{Al}=\dfrac{2.7}{27}=0.1\left(mol\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(0.1........0.3.........0.1...........0.15\)
\(m_{dd_{HCl}}=97.8\cdot1=97.8\left(g\right)\)
\(m_{ddsaupư}=2.7+97.8-0.15\cdot2=100.2\left(g\right)\)
\(C\%_{AlCl_3}=\dfrac{0.1\cdot133.5}{100.2}\cdot100\%=13.32\%\)