a) \(n_C=\dfrac{m}{M}=\dfrac{2,4}{12}=0,2\left(mol\right)\)

\(n_H=\dfrac{m}{M}=\dfrac{0,4}{1}=0,4\left(mol\right)\\ n_O=\dfrac{m}{M}=\dfrac{3,2}{16}=0,2\left(mol\right)\)

\(\Rightarrow\) Trong phân tử của hợp chất có \(1\) nguyên tử C; 2 nguyên tử H; 1 nguyên tử O

\(\Rightarrow CT\) đơn giản : \(\left(CH_2O\right)_n=60\)

\(\Rightarrow30n=60\\ \Rightarrow n=2\\ \Rightarrow CTHH:\left(CH_2O\right)_2\)

b) Gọi \(CTHH\) cần lập là \(Cu_xS_yO_z\)

Theo bài ra, ta có tỉ lệ: \(64x:32y:16z=2:1:2\)

\(\Rightarrow x:y:z=\dfrac{2}{64}:\dfrac{1}{32}:\dfrac{2}{16}\\ \Rightarrow x:y:z=0,03:0,03:0,12\\ \Rightarrow x:y:z=1:1:4\)

\(\Rightarrow CTHH\) cần lập là \(CuSO_4\)

a) \(pthh:2H_2+O_2\overset{t^0}{\rightarrow}2H_2O\left(1\right)\)

b) \(n_{H_2}=\dfrac{V}{22,4}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

Theo \(pthh\left(1\right):n_{O_2}=\dfrac{1}{2}n_{H_2}=\dfrac{1}{2}\cdot0,25=0,125\left(mol\right)\)

\(\Rightarrow V_{O_2}=n\cdot22,4=0,125\cdot22,4=2,8\left(l\right)\\ \Rightarrow V_{k^2}=5V_{O_2}=5\cdot2,8=14\left(l\right)\)

c) \(pthh:2KMnO_4\overset{t^0}{\rightarrow}K_2MnO_4+MnO_2+O_2\left(2\right)\)

Theo \(pthh\left(2\right):n_{KMnO_4}=2n_{O_2}=2\cdot0,125=0,25\left(mol\right)\)

\(\Rightarrow m_{KMnO_4}=n\cdot M=0,25\cdot158=39,5\left(g\right)\)

\(n_{H_2}+n_{C_4H_{10}}=\dfrac{V}{22,4}=\dfrac{17,92}{22,4}=0,8\left(mol\right)\)

Gọi số mol \(C_4H_{10}\) là \(x\left(mol\right)\)

Số mol \(H_2\) là \(3x\left(mol\right)\)

Ta có : \(3x+x=0,8\)

\(\Rightarrow4x=0,8\\ \Rightarrow x=0,2\\ \Rightarrow n_{C_4H_{10}}=0,2\left(mol\right)\\ \Rightarrow n_{H_2}=3\cdot0,2=0,6\left(mol\right)\\ \Rightarrow V_{C_4H_{10}}=n\cdot22,4=0,2\cdot22,4=4,48\left(l\right)\\ V_{H_2}=n\cdot22,4=0,6\cdot22,4=13,44\left(l\right)\\ \Rightarrow\%C_4H_{10}=\dfrac{4,48}{17,92}\cdot100=25\%\\ \%H_2=\dfrac{13,44}{17,92}\cdot100=75\%\)

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96543210 nhéĐặng Trần Minh Anh

a) \(n_{HgO}=\dfrac{m}{M}=\dfrac{43,4}{217}=0,2\left(mol\right)\)

\(pthh:HgO+H_2\underrightarrow{t^0}Hg+H_2O\left(1\right)\)

Theo \(pthh\left(1\right):n_{Hg}=N_{HgO}=0,2\left(mol\right)\)

\(\Rightarrow m_{Hg}=n\cdot M=0,2\cdot201=40,2\left(mol\right)\)

b) Theo \(pthh\left(1\right):n_{H_2}=n_{HgO}=0,2\left(mol\right)\)

\(\Rightarrow V_{H_2}=n\cdot22,4=0,2\cdot22,4=4,48\left(l\right)\)