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Câu 11:
\(\text {Bảo toàn KL: }m_{Fe}+m_{HCl}=m_{FeCl_2}+m_{H_2}\\ \Rightarrow m_{\text{phản ứng}}=m_{Fe_Cl_2}+m_{H_2}=11,61(g)\)
Câu 12:
\(\text{Bảo toàn KL: }m_{Mg}+m_{O_2}=m_{MgO}\\ \Rightarrow m_{O_2}=m_{MgO}-m_{Mg}=15-9=6(g)\)
Câu 13:
\(\text {Bảo toàn KL: }m_{hh}+m_{O_2}=m_{oxit}\\ \Rightarrow m_{O_2}=18-13,2=4,8(g)\)
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\(a,PTHH:2Al+6HCl\to 2AlCl_3+3H_2\\ b,n_{Al}=\dfrac{5,4}{27}=0,2(mol)\\ \Rightarrow n_{H_2}=\dfrac{3}{2}n_{Al}=0,3(mol)\\ \Rightarrow V_{H_2}=0,3.22,4=6,72(l)\\ c,n_{AlCl_3}=n_{Al}=0,2(mol)\\ \Rightarrow m_{AlCl_3}=0,2.133,5=26,7(g)\)
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\(13.A\\ 24.kí.hiệu?\\ 25.C\\ 26.D\\ 27.C\\ 29.D\\ 30.B\\ 31.B\\ 32.V=1,5.24,79=37,185l\\ \Rightarrow D\\ 33.D\\ 34.n_{O_2}=\dfrac{64}{32}=2mol\\ V_{O_2}=2.24,79=49,58l\\ \Rightarrow B\\ 35.d_{SO_2/Cl_2}=\dfrac{61}{71}\approx0,9\\ \Rightarrow C\\ 36.m_{H_2SO_4}=\dfrac{200.12}{100}=24g\\ \Rightarrow C\\ 36.A\\ 39.n_{HCl}=0,2.1=0,2mol\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{Zn}=n_{H_2}=0,2:2=0,1mol\\ m=m_{Zn}=0,1.65=6,5\\ V=V_{H_2}=0,1.22,4=2,24l\\ \Rightarrow A\)
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Bài 1 :
\(n_{CuO}=\dfrac{4.8}{80}=0.06\left(mol\right)\)
\(CuO+H_2\underrightarrow{^{^{t^0}}}Cu+H_2O\)
\(0.06.....0.06...0.06\)
\(V_{H_2}=0.06\cdot22.4=1.344\left(l\right)\)
\(m_{Cu}=0.06\cdot64=3.84\left(g\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(0.06...................................0.06\)
\(m_{Fe}=0.06\cdot56=3.36\left(g\right)\)
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\(D:X_2O_n\\ \%m_X=0,7241=\dfrac{2X}{2X+16n}\\ n=6;X=128\left(Te,Tellurium\right)\\ \Rightarrow D:TeO_3\)
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a.\(n_{ZnO}=\dfrac{20,25}{81}=0,25mol\)
\(ZnO+H_2\rightarrow\left(t^o\right)Zn+H_2O\)
0,25 0,25 ( mol )
\(m_{Zn}=0,25.65=16,25g\)
b.\(n_{H_2SO_4}=\dfrac{49}{98}=0,5mol\)
\(Zn+H_2SO_4\left(l\right)\rightarrow ZnSO_4+H_2\)
0,25 < 0,5 ( mol )
0,25 0,25 ( mol )
\(V_{H_2}=0,25.22,4=5,6l\)
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Câu 1:
Cacbon: C
Sắt: Fe
Oxi: O2
Magie: Mg
Photpho : P
Canxi: Ca
Hidro: H2
Natri: Na
Lưu huỳnh: S
Clo: Cl2
\(\%O=100\%-80\%=20\%\)
Gọi công thức của hợp chất là \(M_xO_y\)
Ta có:
\(n_M:n_O=\frac{80}{M}:\frac{20}{16}=x:y=\frac{80}{M}:1,25\Rightarrow\frac{x}{y}=\frac{64}{M}\Rightarrow32\frac{2y}{x}=M\)
Biên luận ta thấy:
\(2y/x=2;M=64\) là thoả mãn
\(\rightarrow CTHH\) của hợp chất là \(CuO\)