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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
\(d_{O_2;kk}=\frac{M_{O_2}}{M_{kk}}=\frac{2\times16}{29}=\frac{32}{29}\)
a, khối lượng của 2,5 mol CuO là:
\(m=n.M=2,5.80=200\left(g\right)\)
b, số mol của 4,48 lít khí CO2 (đktc) là:
\(n=\dfrac{V}{22,4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
\(a,n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
\(n_{CO_2}=0,5.44=11\left(g\right)\)
\(b,n_{NH_3}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\)
\(n_{NH_3}=0,45,17=7,65\left(g\right)\)
\(c,n_{NO_2}=\dfrac{16,8}{22,4}=0,75\left(mol\right)\)
\(n_{NO_2}=0,75.46=34,5\left(g\right)\)
\(a,n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\ m_{CO_2}=0,25\cdot44=11\left(g\right)\\ b,n_{NH_3}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\\ m_{NH_3}=0,45\cdot17=7,65\left(g\right)\\ c,n_{NO_2}=\dfrac{16,8}{22,4}=0,75\left(mol\right)\\ m_{NO_2}=0,75\cdot46=34,5\left(g\right)\)
\(n_{Al}=\dfrac{13,5}{27}=0,5mol\)
\(n_{H_2SO_4}=\dfrac{98}{98}=1mol\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
0,5 1 0 0
0,5 0,75 0,25 0,75
0 0,25 0,25 0,75
\(V_{H_2}=0,75\cdot22,4=16,8l\)
Chọn B
\(n_{SO_2}=\dfrac{4,958}{24,79}=0,2(mol)\\ \Rightarrow m_{SO_2}=0,2.64=12,8(g)\)
\(a.n_{Na}=\dfrac{4,6}{23}=0,2\left(mol\right)\\ n_{KOH}=\dfrac{8,4}{56}=0,15\left(mol\right)\\ n_{H_3PO_4}=\dfrac{11,76}{98}=0,12\left(mol\right)\\ n_{O_2}=\dfrac{16}{32}=0,5\left(mol\right)\\ b.n_{C_2H_4}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ n_{CO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ n_{N_2}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\)
\(B1\\ n_{O_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\\ 2KClO_3\rightarrow\left(t^o\right)2KCl+3O_2\\ n_{KClO_3}=\dfrac{2}{3}.0,5=\dfrac{1}{3}\left(mol\right)\\ m_{KClO_3}=122,5.\dfrac{1}{3}=\dfrac{245}{6}\left(g\right)\\ B2:n_{O_2}=\dfrac{16}{32}=0,5\left(mol\right)=n_{O_2\left(bài1\right)}\\ \Rightarrow n_{KClO_3}=\dfrac{1}{3}\left(mol\right)\\ m_{KClO_3}=\dfrac{245}{6}\left(g\right)\)
\(a.\)
\(b.\)
\(\Rightarrow m_{SO2}=0,25\times64=16\left(gam\right)\)
\(\Rightarrow m_{H2}=1\times2=2\left(gam\right)\)
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) \(n_{SO_2}=\frac{V}{22,4}=\frac{5,6}{22,4}=0,25\left(mol\right)\)
\(\Rightarrow m_{SO_2}=M.n=64.0,25=16\left(g\right)\)
* \(n_{H_2}=\frac{V}{22,4}=\frac{22,4}{22,4}=1\left(mol\right)\)
\(\Rightarrow m_{H_{ }_2}=M.n=2.1=2\left(g\right)\)