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![](https://rs.olm.vn/images/avt/0.png?1311)
4.
a) \(V_{SO_2}=0.5\cdot22.4=11.2\left(l\right)\)
b) \(V_{CH_4}=\dfrac{3.2}{16}\cdot22.4=4.48\left(l\right)\)
c) \(V_{N_2}=\dfrac{0.9\cdot10^{23}}{6\cdot10^{23}}\cdot22.4=3.36\left(l\right)\)
5.
a) \(m_{Al}=0.1\cdot27=2.7\left(g\right)\)
b) \(m_{Cu\left(NO_3\right)_2}=0.3\cdot188=56.4\left(g\right)\)
c) \(m_{Na_2CO_3}=\dfrac{1.2\cdot10^{23}}{6\cdot10^{23}}\cdot106=21.2\left(g\right)\)
d) \(m_{CO_2}=\dfrac{8.96}{22.4}\cdot44=17.6\left(g\right)\)
e) \(m_K=0.5\cdot2\cdot39=39\left(g\right)\\ m_C=0.5\cdot12=6\left(g\right)\\ m_O=0.5\cdot3\cdot16=24\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(n_{H_2}=\dfrac{1,2.10^{23}}{6.10^{23}}=0,2\left(mol\right)\); \(n_{SO_2}=\dfrac{6,4}{64}=0,1\left(mol\right)\)
V = (1,5 + 2,5 + 0,2 + 0,1).22,4 = 96,32 (l)
b) \(m_{hh}=1,5.32+2,5.28+0,2.2+6,4=124,8\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a. \(m_{CuO}=0.2\cdot80=40\left(g\right)\)
b. \(n_{CO_2}=\dfrac{4.958}{24.79}=0.2\left(mol\right)\Rightarrow m_{CO_2}=0.2\cdot44=8.8\left(g\right)\)
c. \(n_{N_2}=\dfrac{3.011}{6\cdot10^{23}}=5.018\cdot10^{-24}\left(mol\right)\Rightarrow m_{N_2}=5.018\cdot10^{-24}\cdot28=1.405\cdot10^{-22}\left(g\right)\)
Đề câu c có thể nhầm lẫn, bạn kiểm tra lại nha
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(V_{N_2\left(kk\right)}=28.\dfrac{4}{5}=22,4\left(l\right)\)
\(\Rightarrow V_{N_2\left(A\right)}=22,4+11,2=33,6\left(l\right)\)
=> \(\%V_{N2\left(A\right)}=\dfrac{33,6}{5,6+11,2+28}.100=75\%\)
b) Ta có % về thể tích cũng là % về số mol
=> %nN2(A) = 75%
c)\(n_{N_2\left(A\right)}=\dfrac{33,6}{22,4}=1,5\left(mol\right)\)
\(\Rightarrow\%m_{N_2\left(A\right)}=\dfrac{1,5.28}{\dfrac{5,6}{22,4}.17+\dfrac{11,2}{22,4}.28+\dfrac{28}{22,4}.29}.100=77,06\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
- \(V_{CO}=n.24=0,2.24=4,8\left(l\right)\)
- \(n_{SO_3}=\dfrac{m}{M}=\dfrac{8}{80}=0,1\left(mol\right)\)
`=>` \(V_{SO_3}=n.24=0,1.24=2,4\left(l\right)\)
- \(n_{N_2}=\dfrac{\text{Số phân tử}}{6.10^{23}}=\dfrac{3.10^{23}}{6.10^{23}}=0,5\left(mol\right)\)
`=>` \(V_{N_2}=n.24=0,5.24=12\left(l\right)\)
b)
- \(m_{Fe_2O_3}=n.M=0,25.160=40\left(g\right)\)
- \(m_{Al_2O_3}=n.M=0,15.102=15,3\left(g\right)\)
- \(n_{O_2}=\dfrac{V_{\left(\text{đ}ktc\right)}}{22,4}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
`=>` \(m_{O_2}=n.M=0,15.32=4,8\left(g\right)\)
c)
Ta có: \(\left\{{}\begin{matrix}n_{SO_2}=\dfrac{m}{M}=\dfrac{8}{64}=0,125\left(mol\right)\\n_{CO_2}=\dfrac{m}{M}=\dfrac{4,4}{44}=0,1\left(mol\right)\\n_{H_2}=\dfrac{m}{M}=\dfrac{0,1}{2}=0,05\left(mol\right)\end{matrix}\right.\)
`=>` \(n_{hh}=n_{SO_2}+n_{CO_2}+n_{H_2}=0,125+0,1+0,05=0,275\left(mol\right)\)
`=>` \(V_{hh\left(\text{đ}ktc\right)}=n_{hh}.22,4=0,275.22,4=6,16\left(l\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a.
\(m_{Al}=0.5\cdot27=13.5\left(g\right)\)
\(m_{CO_2}=\dfrac{6.72}{22.4}\cdot44=13.2\left(g\right)\)
\(m_{N_2}=\dfrac{5.6}{22.4}\cdot28=7\left(g\right)\)
\(m_{CaCO_3}=0.25\cdot100=25\left(g\right)\)
b.
\(m_{hh}=\dfrac{3.36}{22.4}\cdot2+\dfrac{5.6}{22.4}\cdot28+0.2\cdot44=16.1\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2:
a: \(V=0.2\cdot22.4=4.48\left(lít\right)\)
b: \(n_{N_3}=\dfrac{14}{42}=\dfrac{1}{3}\left(mol\right)\)
\(V=\dfrac{1}{3}\cdot22.4=\dfrac{224}{30}\left(lít\right)\)
3:
a: \(m_{CaCO_3}=0.5\cdot\left(40+12+16\cdot3\right)=50\left(g\right)\)
b: \(n_{SO_2}=\dfrac{5.6}{22.4}=0.25\left(mol\right)\)
\(m_{SO_2}=0.25\cdot\left(32+16\cdot2\right)=16\left(g\right)\)
\(a,m_{Fe}=0,4.56=22,4(g)\\ b,m_{Al_2O_3}=0,025.102=2,55(g)\\ c,m_{SO_3}=\dfrac{80.4,958}{24,79}=16(g)\\ d,m_{N_2}=\dfrac{28.7,437}{24,79}=8,4(g)\\ e,m_{Ag}=\dfrac{108.9.10^{-22}}{6.10^{-23}}=1620(g)\\ f,m_{N_2}=\dfrac{28.1,2.10^{-23}}{6.10^{-23}}=5,6(g)\)