Giải thích các bước giải:
Bài 1:
\(\begin{array}{l} PTHH:S+O_2\xrightarrow{t^o} SO_2\\ Theo\ pt:\ n_{S}=n_{O_2}=n_{SO_2}=0,2\ mol.\\ ⇒m_{S}=0,2\times 32=6,4\ g.\\ ⇒V_{O_2}=0,2\times 22,4=4,48\ lít.\end{array}\)
Bài 2:
\(\begin{array}{l} PTHH:C+O_2\xrightarrow{t^o} CO_2\\ Theo\ pt:\ n_{C}=n_{O_2}=n_{CO_2}=0,18\ mol.\\ ⇒m_{C}=0,18\times 12=2,16\ g.\\ V_{O_2}=0,18\times 22,4=4,032\ lít.\end{array}\)
Bài 3:
\(\begin{array}{l} PTHH:4Al+3O_2\xrightarrow{t^o} 2Al_2O_3\\ Theo\ pt:\ n_{Al}=2n_{Al_2O_3}=0,36\ mol.\\ ⇒m_{Al}=0,36\times 27=9,72\ g.\\ Theo\ pt:\ n_{O_2}=\dfrac{3}{2}n_{Al_2O_3}=0,27\ mol.\\ ⇒V_{O_2}=0,27\times 22,4=6,048\ lít.\end{array}\)
Bài 4:
\(\begin{array}{l} PTHH:4Fe+3O_2\xrightarrow{t^o} 2Fe_2O_3\\ Theo\ pt:\ n_{Fe}=2n_{Fe_2O_3}=0,48\ mol.\\ ⇒m_{Fe}=0,48\times 56=26,88\ g.\\ Theo\ pt:\ n_{O_2}=\dfrac{3}{2}n_{Fe_2O_3}=0,36\ mol.\\ ⇒V_{O_2}=0,36\times 22,4=8,064\ lít.\end{array}\)
Bài 5:
\(\begin{array}{l} PTHH:2Cu+O_2\xrightarrow{t^o} 2CuO\\ Theo\ pt:\ n_{Cu}=n_{CuO}=0,12\ mol.\\ ⇒m_{Cu}=0,12\times 64=7,68\ g.\\ Theo\ pt:\ n_{O_2}=\dfrac{1}{2}n_{CuO}=0,06\ mol.\\ ⇒V_{O_2}=0,06\times 22,4=1,344\ lít.\end{array}\)
Bài 6:
\(\begin{array}{l} PTHH:2Zn+O_2\xrightarrow{t^o} 2ZnO\\ Theo\ pt:\ n_{Zn}=n_{ZnO}=0,12\ mol.\\ ⇒m_{Zn}=0,12\times 65=7,8\ g.\\ Theo\ pt:\ n_{O_2}=\dfrac{1}{2}n_{ZnO}=0,06\ mol.\\ ⇒V_{O_2}=0,06\times 22,4=1,344\ lít.\end{array}\)
Bài 7:
\(\begin{array}{l} PTHH:2Zn+O_2\xrightarrow{t^o} 2ZnO\\ n_{Zn}=\dfrac{6,5}{65}=0,1\ mol.\\ Theo\ pt:\ n_{ZnO}=n_{Zn}=0,1\ mol.\\ ⇒m_{ZnO}=0,1\times 81=8,1\ g.\\ Theo\ pt:\ n_{O_2}=\dfrac{1}{2}n_{Zn}=0,05\ mol.\\ ⇒V_{O_2}=0,05\times 22,4=1,12\ lít.\end{array}\)
Bài 8:
\(\begin{array}{l} PTHH:4K+O_2\xrightarrow{t^o} 2K_2O\\ n_{K}=\dfrac{3,9}{39}=0,1\ mol.\\ Theo\ pt:\ n_{K_2O}=\dfrac{1}{2}n_{K}=0,05\ mol.\\ ⇒m_{K_2O}=0,05\times 94=4,7\ g.\\ Theo\ pt:\ n_{O_2}=\dfrac{1}{4}n_{K}=0,025\ mol.\\ ⇒V_{O_2}=0,025\times 22,4=0,56\ lít.\end{array}\)
Bài 9:
\(\begin{array}{l} PTHH:4Na+O_2\xrightarrow{t^o} 2Na_2O\\ n_{Na}=\dfrac{4,6}{23}=0,2\ mol.\\ Theo\ pt:\ n_{Na_2O}=\dfrac{1}{2}n_{Na}=0,1\ mol.\\ ⇒m_{Na_2O}=0,1\times 62=6,2\ g.\\ Theo\ pt:\ n_{O_2}=\dfrac{1}{4}n_{Na}=0,05\ mol.\\ ⇒V_{O_2}=0,05\times 22,4=1,12\ lít.\end{array}\)
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