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dạ em cảm ơn anh/thầy nhưng mà cái tổng HCl ra m bấm máy sai rồi ạ vs cảm ơn anh/thầy giúp em giải bài nha
Ca+ 2H2O -> Ca(OH)2+ H2
nH2= nCa= 0,14 mol
=> mCa= 5,6g
=> mFe= 6,2-5,6= 0,6g
H2 + O -> H2O
=> Y có 0,14 mol O
nFe2O3= 0,02 mol
=> 0,02 mol Fe2O3 có 0,04 mol Fe và 0,06 mol O
Tổng mol Fe sau phản ứng là \(\dfrac{5,6}{56}\)= 0,1 mol
=> FexOy có 0,06 mol Fe và 0,08 mol O
nFe : nO= 0,06 : 0,08= 3 : 4
=> FexOy là Fe3O4
a= 0,06.56+ 0,08.16= 4,64g
\(n_{H_2}=\dfrac{3,136}{22,4}=0,14mol\)
Gọi \(\left\{{}\begin{matrix}n_{Ca}=x\\n_{Na}=y\end{matrix}\right.\)
\(Ca+2H_2O\rightarrow Ca\left(OH\right)_2+H_2\)
x x ( mol )
\(2Na+2H_2O\rightarrow2NaOH+H_2\)
y 1/2 y ( mol )
Ta có:
\(\left\{{}\begin{matrix}40x+23y=6,2\\x+\dfrac{1}{2}y=0,14\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,04\\y=0,2mol\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}m_{Ca}=0,04.40=1,6g\\m_{Na}=0,2.23=4,6g\end{matrix}\right.\)
\(n_{Fe_2O_3}=\dfrac{3,2}{160}=0,02mol\)
\(Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\)
0,02 0,06 0,04 0,06 ( mol )
\(m_{Fe}=0,04.56=2,24g\)
\(\rightarrow m_{H_2\left(tdFe_xO_y\right)}=0,14-0,06=0,08mol\)
\(n_{Fe\left(tdFe_xO_y\right)}=\dfrac{5,6-0,04.56}{56}=0,06mol\)
\(Fe_xO_y+yH_2\rightarrow\left(t^o\right)xFe+yH_2O\)
0,08 0,06 ( mol )
\(\Rightarrow x:y=0,06:0,08=3:4\)
\(\Rightarrow CTHH:Fe_3O_4\)
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 ...
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)
Bài 1: Ta có: \(n_{H_2}=\dfrac{1,008}{22,4}=0,045\left(mol\right)\)
PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
__0,045__0,09____0,045___0,045 (mol)
a, Ta có: \(a=m_{Mg}=0,045.24=1,08\left(g\right)\)
b, \(V_{ddHCl}=\dfrac{0,09}{0,1}=0,9\left(l\right)\)
c, \(C_{M_{MgCl_2}}=\dfrac{0,045}{0,9}=0,05M\)
Bài 2:
PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
a, Giả sử: \(\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\)
⇒ 24x + 56y = 5,2 (1)
Ta có: \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
Theo PT: \(n_{H_2}=n_{Mg}+n_{Fe}=x+y\left(mol\right)\)
⇒ x + y = 0,15 (2)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,1\left(mol\right)\\y=0,05\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,1.24}{5,2}.100\%\approx46,2\%\\\%m_{Fe}\approx53,8\%\end{matrix}\right.\)
b, Ta có: \(n_{HCl}=2n_{H_2}=0,3\left(mol\right)\)
\(\Rightarrow V_{HCl}=\dfrac{0,3}{1}=0,3\left(l\right)\)
Bạn tham khảo nhé!
\(n_{CO2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
a) Pt : \(MgCO_3+2HCl\rightarrow MgCl_2+CO_2+H_2O|\)
1 2 1 1 1
0,1 0,2 0,1 0,1
\(MgO+2HCl\rightarrow MgCl_2+H_2O|\)
1 2 1 1
0,05 0,05
b) \(n_{HCl}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)
⇒ \(m_{HCl}=0,2.36,5=7,3\left(g\right)\)
\(m_{dd}=\dfrac{7,3.100}{7,3}=100\left(g\right)\)
d) \(n_{MgCO3}=\dfrac{0,2.1}{2}=0,1\left(mol\right)\)
⇒ \(m_{MgCO3}=0,1.84=8,4\left(g\right)\)
\(m_{MgO}=10,4-8,4=2\left(g\right)\)
\(n_{MgO}=\dfrac{2}{40}=0,05\left(mol\right)\)
\(n_{MgCl2}=0,1+0,05=0,15\left(mol\right)\)
⇒ \(m_{MgCl2}=0,15.95=14,25\left(g\right)\)
Sau phản ứng :
\(m_{dd}=10,4+100-\left(0,1.44\right)\)
= 106 (g)
\(C_{MgCl2}=\dfrac{14,25.100}{106}=13,44\)0/0
Chúc bạn học tốt
Ta có: \(n_{H_2}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\)
PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
Theo PT: \(n_{HCl\left(1\right)}=2n_{H_2}=0,7\left(mol\right)\)
Theo ĐLBT KL: mA + mHCl = m muối do hh A sinh ra + mH2
⇒ m muối do hh A sinh ra = 10,7 + 0,7.36,5 - 0,35.2 = 35,55 (g)
Có: \(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)
PT: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)
____0,1____________0,1 (mol)
⇒ c = m muối do hh A sinh ra + mCuCl2 = 35,55 + 0,1.135 = 49,05 (g)
Bạn tham khảo nhé!
Zn +2 HCl ---> ZnCl2 + H2
0,1-----0,2----------0,1-------------0,1 mol
ZnO + 2HCl ---> ZnCl2 + H2O
0,2------0,4-------0,2--------0,2
n H2=\(\dfrac{2,24}{22,4}=0,1mol\)
=>m Zn=0,1.65=6,5g
=>m HCl(1)=0,2.36,5=7,3g
=>m HCl(2)=14,6g -> nHCl=0,4 mol
=>%m Zn=\(\dfrac{6,5}{6,5+14,4}.100=31,1\%\)
=>%m ZnO=68,9%
b)
->m HCl=0,6.36,5=21,9g
->m ZnCl2=0,3.136=40,8g
\(n_{H_2}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(0.1........0.2..................0.1\)
\(n_{CuO}=\dfrac{13.6-0.1\cdot56}{80}=0.1\left(mol\right)\)
\(CuO+2HCl\rightarrow CuCl_2+H_2O\)
\(0.1.......0.2\)
\(C_{M_{HCl}}=\dfrac{0.2+0.2}{0.4}=1\left(M\right)\)
$Fe + 2HCl \to FeCl_2 + H_2$
$CuO + 2HCl \to CuCl_2 +H_2O$
Theo PTHH :
n Fe = n H2 = 2,24/22,4 = 0,1(mol)
=> n CuO = (13,6 - 0,1.56)/80 = 0,1(mol)
n HCl = 2n Fe + 2n CuO = 0,1.2 + 0,1.2 = 0,4(mol)
=> a = CM HCl = 0,4/0,4 = 1(M)