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a) Đặt n CaCO3 =1 ( mol )
=> \(\left\{{}\begin{matrix}n_{Ca}=1\left(mol\right)\\n_C=1\left(mol\right)\\n_O=3\left(mol\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m_{Ca}=40\left(g\right)\\m_C=12\left(g\right)\\m_O=48\left(g\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\%m_{Ca}=40\%\\\%m_C=12\%\\\%m_O=48\%\end{matrix}\right.\)
b) Đặt n H2SO4 = 1 ( mol )
\(\Rightarrow\left\{{}\begin{matrix}n_H=2\left(mol\right)\\n_S=1\left(mol\right)\\n_O=4\left(mol\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m_H=2\left(g\right)\\m_S=32\left(g\right)\\m_O=64\left(g\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\%m_H=2,04\%\\\%m_S=32,65\%\\\%m_O=65,31\%\end{matrix}\right.\)
c) Đặt n Al2S3 = 1 ( mol )
\(\Rightarrow\left\{{}\begin{matrix}n_{Al}=2\left(mol\right)\\n_S=3\left(mol\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m_{Al}=54\left(g\right)\\m_S=96\left(g\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=36\%\\\%m_S=64\%\end{matrix}\right.\)
d) Đặt n CuO = 1 ( mol )
\(\Rightarrow\left\{{}\begin{matrix}n_{Cu}=1\left(mol\right)\\n_O=1\left(mol\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m_{Cu}=64\left(g\right)\\m_O=16\left(g\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\%m_{Cu}=80\%\\\%m_O=20\%\end{matrix}\right.\)
e) Đặt n Fe2(SO4)3 = 1 ( mol )
\(\Rightarrow\left\{{}\begin{matrix}n_{Fe}=2\left(mol\right)\\n_S=3\left(mol\right)\\n_O=12\left(mol\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m_{Fe}=112\left(g\right)\\m_S=96\left(g\right)\\m_O=192\left(g\right)\end{matrix}\right.\Rightarrow}\left\{{}\begin{matrix}\%m_{Fe}=28\%\\\%m_S=24\%\\\%m_O=48\%\end{matrix}\right.\)
e) Đặt n Fe2(SO4)3 = 1 ( mol )
\(\Rightarrow\left\{{}\begin{matrix}n_{Fe}=2\left(mol\right)\\n_S=3\left(mol\right)\\n_O=12\left(mol\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m_{Fe}=112\left(g\right)\\m_S=96\left(g\right)\\m_O=192\left(g\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=28\%\\\%m_S=24\%\\\%m_O=48\%\end{matrix}\right.\)
\(a) n_{Zn(NO_3)_2} = \dfrac{37,8}{189} = 0,2(mol)\\ n_{Zn} = 0,2\ mol \to m_{Zn} = 0,2.65 = 13\ gam\\ n_N = 0,2.2 = 0,4\ mol \to m_N = 0,4.14 = 5,6\ gam\\ m_O = 37,5 - 13 - 5,6 = 18,9(gam)\\ b)n_{Fe_3(PO_4)_2} = \dfrac{10,74}{358} = 0,03(moL)\\ n_{Fe} = 0,03.3 = 0,09 \to m_{Fe} = 0,09.56 = 5,04(gam)\\ n_P = 0,03.2 = 0,06 \to m_P = 0,06.31 = 1,86(gam)\\ m_O = 10,74 - 5,04 -1,86 = 3,84(gam)\\ c) n_{Al} = 0,2.2 = 0,4(mol\to m_{Al} = 0,4.27 = 10,8(gam)\\ n_S = 0,2.3 = 0,6 \to m_S = 0,6.32 = 19,2(gam)\\ n_O = 0,2.12 = 2,4 \to m_O = 2,4.16 = 38,4(gam)\)
\(d) n_{Zn(NO_3)_2} = \dfrac{6.10^{20}}{6.10^{23}} = 0,001(mol)\\ n_{Zn} = 0,001 \to m_{Zn} = 0,001.65 = 0,065(gam)\\ n_N = 0,001.2 = 0,002 \to m_N = 0,002.14 = 0,028(gam)\\ n_O = 0,001.6 = 0,006 \to m_O = 0,006.16= 0,096(gam)\)
Theo gt ta có: $n_{Zn(NO_3)_2}=0,2(mol);n_{Fe_3(PO_4)_2}=0,03(mol);n_{Zn(NO_3)_2}=1(mol)$
a, $m_{Zn}=13(g);m_{N}=5,6(g);m_{O}=19,2(g)$
b, $m_{Fe}=5,04(g);m_{P}=1,86(g)$;m_{O}=3,84(g)$
c, $m_{Al}=10,8(g);m_{S}=19,2(g);m_{O}=38,4(g)$
d, $m_{Zn}=65(g);m_{N}=28(g);m_{O}=96(g)$
Câu a.
\(M_{Ca\left(NO_3\right)_2}=164\)g/mol
\(m_{Ca\left(NO_3\right)_2}=0,3\cdot164=49,2g\)
\(\%Ca=\dfrac{40}{164}\cdot100\%=24,39\%\)
\(m_{Ca}=\%Ca\cdot49,2=12g\)
\(\%N=\dfrac{14\cdot2}{164}\cdot100\%=17,07\%\)
\(m_N=\%N\cdot49,2=8,4g\)
\(m_O=49,2-12-8,4=28,8g\)
Các câu sau em làm tương tự nhé!
a)\(n_{Ca\left(NO_3\right)_2}=0,3mol\)
\(n_{Ca}=n_{Ca\left(NO_3\right)_2}=0,3mol\)
\(m_{Ca}=0,3\cdot40=12g\)
\(n_N=2n_{Ca\left(NO_3\right)_2}=2\cdot0,3=0,6mol\)
\(m_N=0,6\cdot14=8,4g\)
\(n_O=6n_{Ca\left(NO_3\right)_2}=6\cdot0,3=1,8mol\)
\(m_O=1,8\cdot16=28,8g\)
b)\(n_O=\dfrac{9,6}{16}=0,6mol\)
Mà \(n_O=12n_{Fe_2\left(SO_4\right)_3}\Rightarrow n_{Fe_2\left(SO_4\right)_3}=\dfrac{0,6}{12}=0,05mol\)
\(\Rightarrow m=20g\)
c)\(n_{CuSO_4}=\dfrac{3,2}{160}=0,02mol\)
\(n_O=4n_{CuSO_4}=0,08mol=n_{H_2}\)
\(V_{H_2}=0,08\cdot22,4=1,792l\)
\(-PTK_{BaSO_4}=137+32+16.4=233\left(đvC\right)\)
\(-PTK_{Fe\left(OH\right)_3}=56+\left(16+1\right).3=107\left(đvC\right)\)
\(-PTK_{Na_2SO_3}=23.2+32+16.3=126\left(đvC\right)\)
\(-PTK_{Al_2\left(SO_4\right)_3}=27.2+\left(32+16.4\right).3=342\left(đvC\right)\)
\(-PTK_{C_{12}H_{22}O_{11}}=12.12+1.22+16.11=342\left(đvC\right)\)
\(-PTK_{Ca\left(NO_3\right)_2}=40+\left(14+16.3\right).2=164\left(đvC\right)\)
- Xác định hóa trị của Cu trong CuCl:
Biết Cl có hóa trị I. Gọi hóa trị của Cu là a, ta có: 1 × a = 1 × I, rút ra a = I.
- Hóa trị của Mn, S, Fe, Cu, N trong các hợp chất còn lại là:
F e 2 ( S O 4 ) 3 (Fe hóa trị III);
C u ( N O 3 ) 2 , (Cu hóa trị II);
N O 2 (N hóa ttrị IV);
F e C l 2 (Fe hóa trị II);
N 2 O 3 (N hóa trị III);
M n S O 4 (Mn hóa trị II);
S O 3 (S hóa trị VI);
H 2 S (S hóa trị II).
