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\(a.\)
- \(n_{Fe}=\frac{11,2}{56}=0,2\left(mol\right)\)
- \(n_{H2SO4}=\frac{19,6}{98}=0,2\left(mol\right)\)
\(b.\)
- \(n_{SO2}=\frac{5,6}{22,4}=0,25\left(mol\right)\)
\(\Rightarrow m_{SO2}=0,25\times64=16\left(gam\right)\)
- \(n_{H2}=\frac{22,4}{22,4}=1\left(mol\right)\)
\(\Rightarrow m_{H2}=1\times2=2\left(gam\right)\)
a) \(n_{Fe}=\frac{m}{M}=\frac{11,2}{56}=0,2\left(mol\right)\)
\(n_{H_2SO_4}=\frac{m}{M}=\frac{19,6}{98}=0,2\left(mol\right)\)
b) \(n_{SO_2}=\frac{V}{22,4}=\frac{5,6}{22,4}=0,25\left(mol\right)\)
\(\Rightarrow m_{SO_2}=M.n=64.0,25=16\left(g\right)\)
* \(n_{H_2}=\frac{V}{22,4}=\frac{22,4}{22,4}=1\left(mol\right)\)
\(\Rightarrow m_{H_{ }_2}=M.n=2.1=2\left(g\right)\)
a. Ta có: \(n_X=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
Gọi x, y lần lượt là số mol của NO và N2
Ta có: \(\overline{M_X}=\dfrac{30x+28y}{x+y}\left(g\right)\)
Mà \(d_{\dfrac{X}{O_2}}=\dfrac{\overline{M_X}}{M_{O_2}}=\dfrac{\overline{M_X}}{32}=0,9\left(lần\right)\)
=> \(\overline{M_X}=28,8\left(g\right)\)
=> \(\dfrac{30x+28y}{x+y}=28,8\left(g\right)\)
<=> \(1,2x-0,8y=0\) (*)
Theo đề, ta có: x + y = 0,25 (**)
Từ (*) và (**), ta có HPT:
\(\left\{{}\begin{matrix}1,2x-0,8y=0\\x+y=0,25\end{matrix}\right.\)
=> x = 0,1, y = 0,15
=> \(V_{NO}=0,1.22,4=2,24\left(lít\right);V_{N_2}=0,15.22,4=3,36\left(lít\right)\)
b.
\(\%_{V_{NO}}=\dfrac{2,24}{2,24+3,36}.100\%=40\%\)
\(\%_{V_{N_2}}=100\%-40\%=60\%\)
c. Ta có: \(m_{NO}=0,1.30=3\left(g\right)\)
\(m_{N_2}=0,15.28=4,2\left(g\right)\)
=> \(\%_{m_{NO}}=\dfrac{3}{3+4,2}.100\%=41,7\%\)
\(\%_{m_{N_2}}=100\%-41,7\%=58,3\%\)
\(a,n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\ m_{CO_2}=0,25\cdot44=11\left(g\right)\\ b,n_{NH_3}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\\ m_{NH_3}=0,45\cdot17=7,65\left(g\right)\\ c,n_{NO_2}=\dfrac{16,8}{22,4}=0,75\left(mol\right)\\ m_{NO_2}=0,75\cdot46=34,5\left(g\right)\)
\(a,n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
\(n_{CO_2}=0,5.44=11\left(g\right)\)
\(b,n_{NH_3}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\)
\(n_{NH_3}=0,45,17=7,65\left(g\right)\)
\(c,n_{NO_2}=\dfrac{16,8}{22,4}=0,75\left(mol\right)\)
\(n_{NO_2}=0,75.46=34,5\left(g\right)\)
1. \(n_{O_2}=\frac{V}{22,4}=\frac{6,72}{22,4}=0,3\left(mol\right)\)
2.
\(n_{CO_2}=\frac{m}{M}=\frac{4,4}{44}=0,1\left(mol\right)\)
\(n_{O_2}=\frac{m}{M}=\frac{3,2}{32}=0,1\left(mol\right)\)
\(V_{HC}=n.22,4=\left(0,1+0,1\right).22,4=4,48\left(l\right)\)
a) đề bài??
b)
\(n_{SO_3\left(dktc\right)}=\dfrac{V}{22,4}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ m_{SO_3}=n\cdot M=0,1\cdot\left(32+16\cdot3\right)=8\left(g\right)\)
\(a.n_{Na}=\dfrac{4,6}{23}=0,2\left(mol\right)\\ n_{KOH}=\dfrac{8,4}{56}=0,15\left(mol\right)\\ n_{H_3PO_4}=\dfrac{11,76}{98}=0,12\left(mol\right)\\ n_{O_2}=\dfrac{16}{32}=0,5\left(mol\right)\\ b.n_{C_2H_4}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ n_{CO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ n_{N_2}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\)
3. a) MO2/MN2 = 32/28 = 8/7
b) MO2/MCO = 32/28 = 8/7
c) MO2/Mkk = 32/29
1 tính khối lượng của
a) 0.5 mol Fe2O3
\(M_{Fe_2O_3}=2\times56+3\times16=160\) (g/mol)
\(m_{Fe_2O_3}=n_{Fe_2O_3}\times M_{Fe_2O_3}=0,5\times112=56\left(g\right)\)
b) 0,15 mol CO2
\(M_{CO_2}=1\times12+2\times16=44\) (g/mol)
\(m_{CO_2}=n_{CO_2}\times M_{CO_2}=0,15\times44=6,6\left(g\right)\)
c) 5,6 lít O2 ( điều kiện tiêu chuẩn )
\(n_{O_2}=\frac{V_{O_2}}{22,4}=\frac{5,6}{22,4}=0,25\left(mol\right)\)
\(M_{O_2}=2\times16=32\) (g/mol)
\(m_{O_2}=n_{O_2}\times M_{O_2}=0,25\times32=8\left(g\right)\)
d) 8,96 lít H2 ( điều kiện tiêu chuẩn)
\(n_{H_2}=\frac{V_{H_2}}{22,4}=\frac{8,96}{22,4}=0,4\left(mol\right)\)
\(M_{H_2}=2\times1=2\) (g/mol)
\(m_{H_2}=n_{H_2}\times M_{H_2}=0,4\times2=0,8\left(g\right)\)
2 tính thể tích ( điều kiện tiêu chuẩn)
a) 0,125 mol Cl2
\(V_{Cl_2}=22,4\times n_{Cl_2}=22,4\times0,125=2,8\left(l\right)\)
b) 2,5 mol CH4
\(V_{CH_4}=22,4\times n_{CH_4}=22,4\times2,5=56\left(l\right)\)
c) 6,4 gam 02
\(M_{O_2}=2\times16=32\) (g/mol)
\(n_{O_2}=\frac{m_{O_2}}{M_{O_2}}=\frac{6,4}{32}=0,2\left(mol\right)\)
\(V_{O_2}=22,4\times n_{O_2}=22,4\times0,2=4,48\left(l\right)\)
d) 5,6 gam N2
\(M_{N_2}=2\times14=28\) (g/mol)
\(n_{N_2}=\frac{m_{N_2}}{M_{N_2}}=\frac{5,6}{28}=0,2\left(mol\right)\)
\(V_{N_2}=22,4\times n_{N_2}=22,4\times0,2=4,48\left(l\right)\)
3 tính tỉ khối của khí O2 so với
a) khí N2
\(d_{O_2;N_2}=\frac{M_{O_2}}{M_{N_2}}=\frac{2\times16}{2\times14}=\frac{8}{7}\)
b) khí CO
\(d_{O_2;CO}=\frac{M_{O_2}}{M_{CO}}=\frac{2\times16}{1\times12+1\times16}=\frac{8}{7}\)
c) không khí
\(d_{O_2;kk}=\frac{M_{O_2}}{M_{kk}}=\frac{2\times16}{29}=\frac{32}{29}\)