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\(a,PTHH:2Al+6HCl\to 2AlCl_3+3H_2\\ b,n_{Al}=\dfrac{5,4}{27}=0,2(mol)\\ \Rightarrow n_{H_2}=\dfrac{3}{2}n_{Al}=0,3(mol)\\ \Rightarrow V_{H_2}=0,3.22,4=6,72(l)\\ c,n_{AlCl_3}=n_{Al}=0,2(mol)\\ \Rightarrow m_{AlCl_3}=0,2.133,5=26,7(g)\)
a) \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
PTHH: Zn + H2SO4 --> ZnSO4 + H2
0,2-------------->0,2----->0,2
=> \(m_{ZnSO_4}=0,2.161=32,2\left(g\right)\)
b) VH2 = 0,2.22,4 = 4,48 (l)
c) \(n_{CuO}=\dfrac{24}{80}=0,3\left(mol\right)\)
PTHH: CuO + H2 --to--> Cu + H2O
Xét tỉ lệ: \(\dfrac{0,3}{1}>\dfrac{0,2}{1}\) => H2 hết, CuO dư
PTHH: CuO + H2 --to--> Cu + H2O
0,2<---0,2------>0,2
=> mrắn sau pư = 24 - 0,2.80 + 0,2.64 = 20,8 (g)
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\
pthh:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
0,2 0,2 0,2
\(m_{ZnSO_4}=0,2.136=17,2\left(g\right)\\
V_{H_2}=0,2.22,4=4,48\left(l\right)\\
n_{CuO}=\dfrac{24}{80}=0,3\left(mol\right)\\
pthh:CuO+H_2\underrightarrow{t^o}Cu+H_2O\\
LTL:\dfrac{0,3}{1}>\dfrac{0,2}{1}\)
=> CuO dư
\(n_{CuO\left(p\text{ư}\right)}=n_{Cu}=n_{H_2}=0,2\left(mol\right)\\
m_{CuO\left(d\right)}=\left(0,3-0,2\right).80=8\left(g\right)\\
m_{Cr}=8+\left(0,2.64\right)=20,8\left(g\right)\)
(a) Xét \(\dfrac{120}{250}.100=48\) => ddbh và có KCl không bị hoà tan hết
\(\left(b\right)m_{KCl\left(tan\right)}=\dfrac{250}{100}.42,6=106,5\left(g\right)\\ \rightarrow m_{KCl\left(ko.tan\right)}=120-106,5=13,5\left(g\right)\)
Câu 19
a) Fe2O3 + 3H2SO4 --> Fe2(SO4)3 + 3H2O
b) \(n_{H_2SO_4}=\dfrac{24,5}{98}=0,25\left(mol\right)\)
PTHH: Fe2O3 + 3H2SO4 --> Fe2(SO4)3 + 3H2O
0,25---->\(\dfrac{0,25}{3}\)
=> \(m_{Fe_2\left(SO_4\right)_3}=\dfrac{0,25}{3}.400=\dfrac{100}{3}\left(g\right)\)
Bài 1 :
\(n_{CuO}=\dfrac{4.8}{80}=0.06\left(mol\right)\)
\(CuO+H_2\underrightarrow{^{^{t^0}}}Cu+H_2O\)
\(0.06.....0.06...0.06\)
\(V_{H_2}=0.06\cdot22.4=1.344\left(l\right)\)
\(m_{Cu}=0.06\cdot64=3.84\left(g\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(0.06...................................0.06\)
\(m_{Fe}=0.06\cdot56=3.36\left(g\right)\)
\(\left\{{}\begin{matrix}\%Mg=\dfrac{24.1}{120}.100\%=20\%\\\%S=\dfrac{32.1}{120}.100\%=26,667\%\\\%O=\dfrac{16.4}{120}.100\%=53,333\%\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\%Al=\dfrac{27.1}{213}.100\%=12,676\%\\\%N=\dfrac{14.3}{213}.100\%=19,718\%\\\%O=\dfrac{16.9}{213}.100\%=67,606\%\end{matrix}\right.\)
\(D:X_2O_n\\ \%m_X=0,7241=\dfrac{2X}{2X+16n}\\ n=6;X=128\left(Te,Tellurium\right)\\ \Rightarrow D:TeO_3\)
Khối lượng AgNO3 có thể tan trong 250g nước ở 25oC là:
\(\dfrac{250.222}{100}=555\left(g\right)\)
Em cảm ơn nhìu lắm:>