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\(m_{HCl}=200.14,6\%=29,2\left(g\right)\)
\(n_{HCl}=\frac{29,2}{36,5}=0,8\left(mol\right)\)
\(Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\)
x__________________2x______x____________
\(KHCO_3+HCl\rightarrow KCl+CO_2+H_2O\)
y__________________y______y____________
\(n_{NaOH}=0,4.1=0,4\left(mol\right)\)
Giải hệ PT:
\(\left\{{}\begin{matrix}106x+100y=25,9\\2x+y=0,4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,15\\y=0,1\end{matrix}\right.\)
\(\Rightarrow n_{CO2}=x=0,15\left(mol\right)\)
a,\(V_{CO2}=0,15.22,4=3,36\left(l\right)\)
b,
\(\%_{NaCl}=\frac{\left(2.0,15\right).58,5}{\left(2.0,15\right).58,5+0,1.74,5}.100\%=70,2\%\)
\(\Rightarrow\%_{KCl}=100\%-70,2\%=29,8\%\)
a, PT: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)
b, Ta có: \(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
Theo PT: \(n_{Zn}=n_{H_2}=0,5\left(mol\right)\)
\(\Rightarrow m_{Zn}=0,5.65=32,5\left(g\right)\)
\(\Rightarrow m_{CuO}=72,5-32,5=40\left(g\right)\)
c, Ta có: \(n_{CuO}=\dfrac{40}{80}=0,5\left(mol\right)\)
Theo PT: \(n_{H_2SO_4}=n_{Zn}+n_{CuO}=1\left(mol\right)\)
\(\Rightarrow b=C_{M_{H_2SO_4}}=\dfrac{1}{2,5}=0,4M\)
c, Theo PT: \(\left\{{}\begin{matrix}n_{ZnSO_4}=n_{Zn}=0,5\left(mol\right)\\n_{CuSO_4}=n_{CuO}=0,5\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}C_{M_{ZnSO_4}}=\dfrac{0,5}{2,5}=0,2M\\C_{M_{CuSO_4}}=\dfrac{0,5}{2,5}=0,2M\end{matrix}\right.\)
Bạn tham khảo nhé!
a, \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\left(I\right)\)
\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\left(II\right)\)
b, Theo PTHH(1) : \(n_{Zn}=n_{H_2}=0,5\left(mol\right)\)
\(\Rightarrow m_{Zn}=32,5\left(g\right)\)
\(\Rightarrow m_{CuO}=m_{hh}-m_{Zn}=40\left(g\right)\)
\(\Rightarrow n_{CuO}=\dfrac{m}{M}=0,5\left(mol\right)\)
c, Theo PTHH (1) và (2) : \(n_{H2SO4}=n_{CuO}+n_{Zn}=1\left(mol\right)\)
\(\Rightarrow C_{MH2SO4}=b=\dfrac{n}{V}=\dfrac{1}{2,5}=0,4M\)
d, ( Chắc là thể tích coi như không đổi )
Thấy sau phản ứng thu được A gồm \(0,5molZnSO_4,0,5molCuSO_4\)
\(\Rightarrow C_{MCuSO4}=C_{MZnSO4}=\dfrac{n}{V}=\dfrac{0,5}{2,5}=0,2M\)
Vậy ...
\(n_{H_2SO_4}=0,25.2=0,5\left(mol\right)\)
Gọi \(\left\{{}\begin{matrix}n_{Al_2O_3}=x\left(mol\right)\\n_{CuO}=y\left(mol\right)\end{matrix}\right.\)
\(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)
x----------> 3x --------> x
\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)
y --------> y --------> y
Có hệ phương trình
\(\left\{{}\begin{matrix}102x+80y=26,2\\3x+y=0,5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)
\(\%_{m_{Al_2O_3}}=\dfrac{102.0,1.100}{26,2}=38,93\%\)
\(\%_{m_{CuO}}=\dfrac{80.0,2.100}{26,2}=61,07\%\)
\(CM_{Al_2\left(SO_4\right)_3}=\dfrac{x}{0,25}=\dfrac{0,1}{0,25}=0,4M\)
\(CM_{CuSO_4}=\dfrac{y}{0,25}=\dfrac{0,2}{0,25}=0,8M\)
a)
$n_{HCl} = \dfrac{250.14,6\%}{36,5} = 1(mol)$
$Na_2CO_3 + 2HCl \to 2NaCl + CO_2 + H_2O$
$n_{CO_2} = \dfrac{1}{2}n_{HCl} = 0,5(mol)$
$V_{CO_2} = 0,5.22,4 = 11,2(lít)$
b)
Sau phản ứng :
$m_{dd} = 55 + 250 -0,5.44 = 283(gam)$
$n_{Na_2CO_3} = n_{CO_2} = 0,5(mol) \Rightarrow m_{Na_2SO_4} = 55 - 0,5.106 = 2(gam)$
$n_{NaCl} =n_{HCl} = 1(mol)$
$C\%_{NaCl} = \dfrac{1.58,5}{283}.100\% = 20,67\%$
$C\%_{Na_2SO_4} = \dfrac{2}{283}.100\% = 0,71\%$
a) \(Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\)
\(n_{HCl}=\dfrac{250.14,6\%}{36,5}=1\left(mol\right)\)
\(TheoPT:n_{CO_2}=\dfrac{1}{2}n_{HCl}=0,5\left(mol\right)\)
=> \(V_{CO_2}=0,5.22,4=11,2\left(l\right)\)
b) \(C\%_{NaCl}=\dfrac{0,5.58,5}{55+250-0,5.44}.100=10,34\%\)
\(m_{Na_2SO_4}=55-0,5.106=2\left(g\right)\)
=> \(C\%_{Na_2SO_4}=\dfrac{2}{55+250-0,5.44}.100=0,7\%\)
Bài 1:
PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
Ta có: \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,4\left(mol\right)\\n_{H_2}=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{ddHCl}=\dfrac{0,4\cdot36,5}{14,6\%}=100\left(g\right)\\V_{H_2}=0,2\cdot22,4=4,48\left(l\right)\end{matrix}\right.\)
Bài 2:
PTHH: \(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)
Ta có: \(\left\{{}\begin{matrix}n_{KOH}=\dfrac{100\cdot11,2\%}{56}=0,2\left(mol\right)\\n_{H_2SO_4}=\dfrac{150\cdot9,8\%}{98}=0,15\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,2}{2}< \dfrac{0,15}{1}\) \(\Rightarrow\) H2SO4 còn dư, KOH p/ứ hết
\(\Rightarrow\left\{{}\begin{matrix}n_{K_2SO_4}=0,1\left(mol\right)\\n_{H_2SO_4\left(dư\right)}=0,05\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{K_2SO_4}=0,1\cdot174=17,4\left(g\right)\\m_{H_2SO_4\left(dư\right)}=0,05\cdot98=4,9\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{ddKOH}+m_{ddH_2SO_4}=250\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{K_2SO_4}=\dfrac{17,4}{250}\cdot100\%=6,96\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{4,9}{250}\cdot100\%=1,96\%\end{matrix}\right.\)
a, PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
b, Gọi: \(\left\{{}\begin{matrix}n_{Fe}=x\left(mol\right)\\n_{Mg}=y\left(mol\right)\end{matrix}\right.\)
Theo PT: \(\left\{{}\begin{matrix}n_{HCl}=2n_{Fe}+2n_{Mg}=2x+2y\left(mol\right)\\n_{H_2}=n_{Fe}+n_{Mg}=x+y\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow m_{HCl}=36,5.\left(2x+2y\right)=73\left(x+y\right)\left(g\right)\)
\(\Rightarrow m_{ddHCl}=\dfrac{73\left(x+y\right)}{20\%}=365\left(x+y\right)\left(g\right)\)
Ta có: m dd sau pư = mFe + mMg + m dd HCl - mH2 = 56x + 24y + 365.(x+y) - 2.(x+y) = 419x + 387y (g)
Theo PT: \(n_{MgCl_2}=n_{Mg}=y\left(mol\right)\)
\(C\%_{MgCl_2}=11,87\%\) \(\Rightarrow\dfrac{95y}{419x+387y}=0,1187\)
\(\Rightarrow\dfrac{x}{y}=0,9865\Rightarrow x=0,9865y\)
Theo PT: \(n_{FeCl_2}=n_{Fe}=x\left(mol\right)\)
\(\Rightarrow C\%_{FeCl_2}=\dfrac{127x}{419x+387y}.100\%=\dfrac{127.0,9865y}{419.0,9865y+387y}.100\%\approx15,65\%\)
Bài 1:
PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
Ta có: \(n_{Mg}=\dfrac{1}{2}n_{HCl}=\dfrac{1}{2}\cdot0,11\cdot1,5=0,0825\left(mol\right)\)
\(\Rightarrow m_{Mg}=0,0825\cdot24=1,98\left(g\right)\)
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
\(n_{Fe}=\dfrac{11.2}{56}=0.2\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(0.2...................0.2..........0.2\)
\(V_{H_2}=0.2\cdot22.4=4.48\left(l\right)\)
\(m_{FeCl_2}=0.2\cdot127=25.4\left(g\right)\)
\(m_{\text{dung dịch sau phản ứng}}=11.2+200-0.2\cdot2=210.8\left(g\right)\)
\(C\%_{FeCl_2}=\dfrac{25.4}{210.8}\cdot100\%=12.05\%\)
a. \(m_{HCl}=200.14,6\%=29,2g\)
\(n_{HCl}=\frac{29,2}{36,5}=0,8mol\)
\(n_{NaOH}=\frac{400}{1000}.1=0,4mol\)
Để trung hoà hết Y cần 0,4mol NaOH, suy ra HCl dư
\(n_{HCl_{pu}}=0,8-0,4=0,4mol\)
PTHH: \(Na_2CO_3+2HCl\rightarrow2NaCl+CO_2\uparrow+H_2O\)
\(KHCO_3+HCl\rightarrow KCl+CO_2\uparrow+H_2O\)
Đặt \(\hept{\begin{cases}x\left(mol\right)=n_{Na_2CO_3}\\y\left(mol\right)=n_{KHCO_3}\end{cases}}\)
\(\rightarrow106x+100y=25,9g\left(1\right)\)
Theo phương trình \(n_{HClpu}=2n_{Na_2CO_3}+n_{KHCO_3}=0,4mol\)
\(\rightarrow2x+y=0,4mol\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\hept{\begin{cases}x=0,15mol\\y=0,1mol\end{cases}}\)
Theo phương trình \(n_{CO_2}=n_{Na_2CO_3}+n_{KHCO_3}=0,15+0,1=0,25mol\)
\(V_{CO_2}=0,25.22,4=5,6l\)
b. \(m_{dd}=25,9+200-0,25.44=214,9g\)
\(C\%_{NaCl}=\frac{2.0,15.58,5}{214,9}.100\%=8,2\%\)
\(C\%_{KCl}=\frac{0,1.74,5}{214,9}.100\%=3,5\%\)