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\(n_{CO2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
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
\(a.n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ a............3a.......a.........1,5a\left(mol\right)\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ b.........2b........b.........b\left(mol\right)\\ \rightarrow\left\{{}\begin{matrix}27a+56b=5,5\\1,5a+b=0,2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,05\end{matrix}\right.\\ \rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{27.0,1}{5,5}.100\approx49,091\%\\\%m_{Fe}=\dfrac{0,05.56}{5,5}.100\approx50,909\%\end{matrix}\right.\\ b.C_{MddHCl}=\dfrac{3a+2b}{0,5}=\dfrac{3.0,1+2.0,05}{0,5}=0,8\left(M\right)\)
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
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)
Mg + 2HCl → MgCl2 + H2 (1)
MgO + 2HCl → MgCl2 + H2O (2)
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
a) Theo PT1: \(n_{Mg}=n_{H_2}=0,1\left(mol\right)\)
\(\Rightarrow m_{Mg}=0,1\times24=2,4\left(g\right)\)
\(\Rightarrow m_{MgO}=4,4-2,4=2\left(g\right)\)
\(\Rightarrow\%Mg=\dfrac{2,4}{4,4}\times100\%=54,55\%\)
\(\%MgO=\dfrac{2}{4,4}\times100\%=45,45\%\)
b) Theo PT1: \(n_{HCl}=2n_{Mg}=2\times0,1=0,2\left(mol\right)\)
\(n_{MgO}=\dfrac{2}{40}=0,05\left(mol\right)\)
Theo PT2: \(n_{HCl}=2n_{MgO}=2\times0,05=0,1\left(mol\right)\)
\(\Rightarrow\Sigma n_{HCl}=0,2+0,1=0,3\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}=\dfrac{0,3}{0,3}=1\left(M\right)\)
c) Theo PT1,2: \(\Sigma n_{MgCl_2}=\dfrac{1}{2}\Sigma n_{HCl}=\dfrac{1}{2}\times0,3=0,15\left(mol\right)\)
\(\Rightarrow C_{M_{MgCl_2}}=\dfrac{0,15}{0,3}=0,5\left(M\right)\)
a,Fe + 2HCl → FeCl + H2 (1)
FeO + 2HCl → FeCl + H2O (2)
nH2 = 3,36/ 22,4 = 0,15 ( mol)
Theo (1) nH2 = nFe = 0,15 ( mol)
mFe = 0,15 x 56 = 8.4 (g)
m FeO = 12 - 8,4 = 3,6 (g)
a, \(n_{H_2}=\frac{3,36}{22,4}=0,15\left(mol\right)\)
\(Fe+2HCl->FeCl_2+H_2\left(1\right)\)
\(FeO+2HCl->FeCl_2+H_2O\left(2\right)\)
theo (1) \(n_{Fe}=n_{H_2}=0,15\left(mol\right)\)
=> \(m_{Fe}=0,15.56=8,4\left(g\right)\)
=> \(m_{FeO}=12-8,4=3,6\left(g\right)\)
ta thấy : nFe =nH2 = 0,15
=> mFe =0,15 x 56 = 8,4g
%Fe=8,4/12 x 100 = 70%
=>%FeO = 100 - 70 = 30%
b) BTKLra mdd tìm mct of HCl
c) tìm mdd sau pứ -mH2 nha bạn
Gọi x,y lần lượt là số mol của Mg, MgO
\(Pt:Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\) (1)
x ----------> 0,4 ----------> x
\(MgO+2HCl\rightarrow MgCl_2+H_2O\) (2)
y --------> 0,2 -------> y
(1)(2) \(\Rightarrow\left\{{}\begin{matrix}24x+40y=8,8\\95x+95y=28,5\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)
\(\%Mg=\dfrac{0,2.24.100}{8,8}=54,54\%\)
\(\%MgO=100-54,54=45,45\%\)
\(m_{dd_{HCl}}=\dfrac{\left(0,4+0,2\right).36,5.100}{7,3}=300\left(g\right)\)
\(n_{H_2}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(0.1........0.2..................0.1\)
\(n_{CuO}=\dfrac{13.6-0.1\cdot56}{80}=0.1\left(mol\right)\)
\(CuO+2HCl\rightarrow CuCl_2+H_2O\)
\(0.1.......0.2\)
\(C_{M_{HCl}}=\dfrac{0.2+0.2}{0.4}=1\left(M\right)\)
MgO + 2HCl → MgCl2 + H2O
Fe + 2HCl → FeCl2 + H2
a. \(n_{H_2}\)= 0,1 mol
⇒ nFe = 0,1 mol ⇒ mFe = 5,6 gam
⇒ mMgO = 40 gam ⇒ nMgO = 0,1 mol
b. Thể tích HCl đã dùng là
V = \(\dfrac{n}{C_M}\) = \(\dfrac{0,1.2+0,1.2}{1}\) = 0,4 lít = 400 mL