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\(a)\)
\(n_C=\frac{24}{12}=2mol\)
\(n_O=\frac{32}{16}=2mol\)
\(\frac{n_C}{n_O}=\frac{2}{2}=\frac{1}{1}\)
\(\rightarrow CTHH:CO\)
\(b)\)
\(n_{Na}=\frac{46}{23}=2\)
\(n_O=\frac{16}{16}=1\)
\(n_{Na}:n_O=2:1\)
\(\rightarrow CTHH:Na_2O\)
\(c)\)
\(n_{Cu}=\frac{32}{64}=0,5mol\)
\(n_S=\frac{16}{32}=0,5mol\)
\(n_O=\frac{32}{16}=2mol\)
\(n_{Cu}:n_S:n_O=0,5:0,5:2=1:1:4\)
\(\rightarrow CTHH:CuSO_4\)
a, PT: \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
Ta có: \(n_{KClO_3}=\dfrac{24,5}{122,5}=0,2\left(mol\right)\)
Theo PT: \(n_{O_2}=\dfrac{3}{2}n_{KClO_3}=0,3\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,3.24,79=7,437\left(g\right)\)
b, PT: \(2Cu+O_2\underrightarrow{t^o}2CuO\)
Ta có: \(n_{Cu}=\dfrac{32}{64}=0,5\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,5}{2}< \dfrac{0,3}{1}\), ta được O2 dư.
Theo PT: \(\left\{{}\begin{matrix}n_{O_2\left(pư\right)}=\dfrac{1}{2}n_{Cu}=0,25\left(mol\right)\\n_{CuO}=n_{Cu}=0,5\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow n_{O_2\left(dư\right)}=0,3-0,25=0,05\left(mol\right)\)
\(\Rightarrow m_{O_2\left(dư\right)}=0,05.32=1,6\left(g\right)\)
\(m_{CuO}=0,5.80=40\left(g\right)\)
a)
\(n_{SO_2\left(dktc\right)}=\dfrac{V}{22,4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ m_{SO_2}=n\cdot M=0,2\cdot\left(32+16\cdot2\right)=12,8\left(g\right)\)
b)
\(n_{CH_4}=\dfrac{m}{M}=\dfrac{6,4}{12+1\cdot4}=0,4\left(mol\right)\\ V_{CH_4\left(dktc\right)}=n\cdot22,4=0,4\cdot22,4=8,96\left(l\right)\)
\(NTK_x=2NTK_O=16\cdot2=32\left(đvC\right)\)
Vậy X là lưu huỳnh (S)
\(n_{Al}=\dfrac{6,75}{27}=0,25mol\)
\(n_{O_2}=\dfrac{3,36}{22,4}=0,15mol\)
\(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
0,25 0,15 0
0,2 0,15 0,1
0,05 0 0,1
\(m_{dư}=m_{Aldư}=0,05\cdot27=1,35g\)
\(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\uparrow\)
0,3 0,15
\(m_{KMnO_4}=0,3\cdot158=47,4g\)
\(\%m_K=\dfrac{39.2}{174}.100\%=44,83\%\)