1)
\(n_S=\dfrac{3.2}{32}=0.1\left(mol\right)\)
\(n_{O_2}=\dfrac{1.12}{22.4}=0.05\left(mol\right)\)
\(S+O_2\underrightarrow{^{^{t^0}}}SO_2\)
Ta có :
\(n_S=0.1>n_{O_2}=0.05\)
=> S dư
\(n_{O_2}=n_{SO_2}=0.05\left(mol\right)\)
\(V_{SO_2}=0.05\cdot22.4=1.12\left(l\right)\)
2)
\(n_{HCl}=0.2\cdot0.5=0.1\left(mol\right)\)
\(A+2HCl\rightarrow ACl_2+H_2\)
\(0.05...0.1....................0.05\)
\(M_A=\dfrac{3.25}{0.05}=65\left(\dfrac{g}{mol}\right)\)
\(A:Zn\)
\(V_{H_2}=0.05\cdot22.4=1.12\left(l\right)\)
Bài 1:
\(PTHH:S+O_2\rightarrow^{t^o}SO_2\\ n_S=\dfrac{3,2}{32}=0,1\left(mol\right);n_{O_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\\ \text{Vì }\dfrac{n_S}{1}>\dfrac{n_{O_2}}{2}\rightarrow S\text{ dư}\\ \Rightarrow n_{SO_2}=0,05\left(mol\right)\\ \Rightarrow V_{SO_2\left(đktc\right)}=0,05\cdot22,4=1,12\left(l\right)\)
Bài 2:
Gọi KL cần tìm là R
\(\Rightarrow n_R=\dfrac{3,25}{M_R}\left(mol\right);n_{HCl}=0,5\cdot0,2=0,1\left(mol\right)\\ PTHH:R+2HCl\rightarrow RCl_2+H_2\\ \Rightarrow n_R=\dfrac{1}{2}n_{HCl}=0,05\left(mol\right)\\ \Rightarrow\dfrac{3,25}{M_R}=0,05\\ \Rightarrow M_R=65\left(g/mol\right)\\ \Rightarrow R\text{ là kẽm }\left(Zn\right)\\ PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ \Rightarrow n_{H_2}=n_{Zn}=0,05\left(mol\right)\\ \Rightarrow V_{H_2\left(đktc\right)}=0,05\cdot22,4=1,12\left(l\right)\)
Bài 3: Sửa: Đốt cháy hoàn toàn 10,8g kim loại M(III)
\(n_M=\dfrac{10,8}{M_M}\left(mol\right);n_{M_2O_3}=\dfrac{20,4}{2M_M+48}\left(mol\right)\\ PTHH:4M+3O_2\rightarrow^{t^o}2M_2O_3\\ \Rightarrow n_M=2n_{M_2O_3}\\ \Rightarrow\dfrac{10,8}{M_M}=\dfrac{40,8}{2M_M+48}\\ \Rightarrow21,6M_M+518,4=40,8M_M\\ \Rightarrow19,2M_M=518,4\\ \Rightarrow M_M=27\left(g/mol\right)\\ \Rightarrow M\text{ là nhôm }\left(Al\right)\\ \Rightarrow n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\\ \Rightarrow n_{O_2}=\dfrac{3}{4}n_{Al}=0,3\left(mol\right)\\ \Rightarrow V_{O_2\left(đktc\right)}=0,3\cdot22,4=6,72\left(l\right)\)
