Đáp án:
Giải thích các bước giải:
\(3/\\ a,PTHH:2C_4H_{10}+13O_2\xrightarrow{t^o} 8CO_2+10H_2O\\ B,n_{O_2}=\dfrac{4,48}{22,4}=0,2\ mol.\\ Theo\ pt:\ n_{C_4H_{10}}=\dfrac{2}{13}n_{O_2}=\dfrac{2}{65}\ mol.\\ ⇒V_{C_4H_{10}}=\dfrac{2}{65}.22,4=\dfrac{224}{325}\ lít.\\ Theo\ pt:\ n_{CO_2}=\dfrac{8}{13}n_{O_2}=\dfrac{8}{65}\ mol.\\ ⇒V_{CO_2}=\dfrac{8}{65}.22,4=\dfrac{896}{325}\ lít.\\ c,PTHH:\\ 2KMnO_4\xrightarrow{t^o} K_2MnO_4+MnO_2+O_2\ (1)\\ 2KClO_3\xrightarrow{t^o} 2KCl+3O_2\ (2)\\ Theo\ pt\ (1):\ n_{KMnO_4}=2n_{O_2}=0,4\ mol.\\ ⇒m_{KMnO_4}=0,4.158=63,2\ g.\\ Theo\ pt\ (2):\ n_{KClO_3}=\dfrac{2}{3}n_{O_2}=\dfrac{2}{15}\ mol.\\ ⇒m_{KClO_3}=\dfrac{2}{15}.122,5=16,33\ g.\\ 4/\\ a,PTHH:\\ 2Cu+O_2\xrightarrow{t^o} 2CuO\ (1)\\ 3Fe+2O_2\xrightarrow{t^o} Fe_3O_4\ (2)\\ 2Ba+O_2\xrightarrow{t^o} 2BaO\ (3)\\ Theo\ pt\ (1),(2),(3):\ n_{O_2}=\dfrac{1}{2}n_{Cu}+\dfrac{2}{3}n_{Fe}+\dfrac{1}{2}n_{Ba}=0,125+0,06+0,375=0,56\ mol.\\ ⇒V_{O_2}=0,56.22,4=12,544\ lít.\\ c,Theo\ pt\ (1):\ n_{CuO}=n_{Cu}=0,25\ mol.\\ ⇒m_{CuO}=0,25.80=20\ g.\\ Theo\ pt\ (2):\ n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=0,03\ mol.\\ ⇒m_{Fe_3O_4}=0,03.232=6,96\ g.\\ Theo\ pt\ (3):\ n_{BaO}=n_{Ba}=0,75\ mol.\\ ⇒m_{BaO}=0,75.153=114,75\ g.\\ ⇒m_{sản\ phẩm}=m_{CuO}+m_{Fe_3O_4}+m_{BaO}=20+6,96+114,75=141,71\ g.\)
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